Proof of exponential property












0












$begingroup$


Given $f(x) = exp(ax)$, then the derivative, by definition is given by $frac{df}{dx} = aexp(ax)$. Also, it is known that $f(0) = 1$



I want to prove that $exp((a+b)x)$ is equal to $exp(ax)cdotexp(bx)$.



So, to do this, I take the derivative of $exp((a+b)x)$ which results in $(a+b)exp((a+b)x)$ by definition of the exponential. Next, I take the derivative of $exp(ax)cdotexp(bx)$ which results in $(a+b)exp(ax)cdotexp(bx)$.



Now, I evaluate $frac{d}{dx}$$[exp((a+b)x) - exp(ax)cdotexp(bx)]$ at $x = 0$. This results in zero.



My question now is, have I completed the proof. If yes, can someone explain why?






calculus exponential-function







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    No. You have two functions $f$ and $g$, that you differentiate, and you check the derivatives are equal at $0$. Does it mean $f$ and $g$ are equal everywhere? Take for instance $f(x)=x^2$ and $g(x)=2x^2$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 6 '18 at 1:22












  • $begingroup$
    So how would I continue?
    $endgroup$
    – GustoCo
    Dec 6 '18 at 1:45










  • $begingroup$
    To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
    $endgroup$
    – AlexanderJ93
    Dec 6 '18 at 3:44










  • $begingroup$
    To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
    $endgroup$
    – AlexanderJ93
    Dec 6 '18 at 3:44
















0












$begingroup$


Given $f(x) = exp(ax)$, then the derivative, by definition is given by $frac{df}{dx} = aexp(ax)$. Also, it is known that $f(0) = 1$



I want to prove that $exp((a+b)x)$ is equal to $exp(ax)cdotexp(bx)$.



So, to do this, I take the derivative of $exp((a+b)x)$ which results in $(a+b)exp((a+b)x)$ by definition of the exponential. Next, I take the derivative of $exp(ax)cdotexp(bx)$ which results in $(a+b)exp(ax)cdotexp(bx)$.



Now, I evaluate $frac{d}{dx}$$[exp((a+b)x) - exp(ax)cdotexp(bx)]$ at $x = 0$. This results in zero.



My question now is, have I completed the proof. If yes, can someone explain why?






calculus exponential-function







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    No. You have two functions $f$ and $g$, that you differentiate, and you check the derivatives are equal at $0$. Does it mean $f$ and $g$ are equal everywhere? Take for instance $f(x)=x^2$ and $g(x)=2x^2$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 6 '18 at 1:22












  • $begingroup$
    So how would I continue?
    $endgroup$
    – GustoCo
    Dec 6 '18 at 1:45










  • $begingroup$
    To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
    $endgroup$
    – AlexanderJ93
    Dec 6 '18 at 3:44










  • $begingroup$
    To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
    $endgroup$
    – AlexanderJ93
    Dec 6 '18 at 3:44














0












0








0





$begingroup$


Given $f(x) = exp(ax)$, then the derivative, by definition is given by $frac{df}{dx} = aexp(ax)$. Also, it is known that $f(0) = 1$



I want to prove that $exp((a+b)x)$ is equal to $exp(ax)cdotexp(bx)$.



So, to do this, I take the derivative of $exp((a+b)x)$ which results in $(a+b)exp((a+b)x)$ by definition of the exponential. Next, I take the derivative of $exp(ax)cdotexp(bx)$ which results in $(a+b)exp(ax)cdotexp(bx)$.



Now, I evaluate $frac{d}{dx}$$[exp((a+b)x) - exp(ax)cdotexp(bx)]$ at $x = 0$. This results in zero.



My question now is, have I completed the proof. If yes, can someone explain why?






calculus exponential-function







share|cite|improve this question











$endgroup$




Given $f(x) = exp(ax)$, then the derivative, by definition is given by $frac{df}{dx} = aexp(ax)$. Also, it is known that $f(0) = 1$



I want to prove that $exp((a+b)x)$ is equal to $exp(ax)cdotexp(bx)$.



So, to do this, I take the derivative of $exp((a+b)x)$ which results in $(a+b)exp((a+b)x)$ by definition of the exponential. Next, I take the derivative of $exp(ax)cdotexp(bx)$ which results in $(a+b)exp(ax)cdotexp(bx)$.



Now, I evaluate $frac{d}{dx}$$[exp((a+b)x) - exp(ax)cdotexp(bx)]$ at $x = 0$. This results in zero.



My question now is, have I completed the proof. If yes, can someone explain why?






calculus exponential-function




calculus exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 5:23









Tianlalu

3,08621038




3,08621038










asked Dec 6 '18 at 1:05









GustoCoGustoCo

428




428








  • 3




    $begingroup$
    No. You have two functions $f$ and $g$, that you differentiate, and you check the derivatives are equal at $0$. Does it mean $f$ and $g$ are equal everywhere? Take for instance $f(x)=x^2$ and $g(x)=2x^2$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 6 '18 at 1:22












  • $begingroup$
    So how would I continue?
    $endgroup$
    – GustoCo
    Dec 6 '18 at 1:45










  • $begingroup$
    To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
    $endgroup$
    – AlexanderJ93
    Dec 6 '18 at 3:44










  • $begingroup$
    To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
    $endgroup$
    – AlexanderJ93
    Dec 6 '18 at 3:44














  • 3




    $begingroup$
    No. You have two functions $f$ and $g$, that you differentiate, and you check the derivatives are equal at $0$. Does it mean $f$ and $g$ are equal everywhere? Take for instance $f(x)=x^2$ and $g(x)=2x^2$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 6 '18 at 1:22












  • $begingroup$
    So how would I continue?
    $endgroup$
    – GustoCo
    Dec 6 '18 at 1:45










  • $begingroup$
    To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
    $endgroup$
    – AlexanderJ93
    Dec 6 '18 at 3:44










  • $begingroup$
    To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
    $endgroup$
    – AlexanderJ93
    Dec 6 '18 at 3:44








3




3




$begingroup$
No. You have two functions $f$ and $g$, that you differentiate, and you check the derivatives are equal at $0$. Does it mean $f$ and $g$ are equal everywhere? Take for instance $f(x)=x^2$ and $g(x)=2x^2$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 1:22






$begingroup$
No. You have two functions $f$ and $g$, that you differentiate, and you check the derivatives are equal at $0$. Does it mean $f$ and $g$ are equal everywhere? Take for instance $f(x)=x^2$ and $g(x)=2x^2$.
$endgroup$
– Jean-Claude Arbaut
Dec 6 '18 at 1:22














$begingroup$
So how would I continue?
$endgroup$
– GustoCo
Dec 6 '18 at 1:45




$begingroup$
So how would I continue?
$endgroup$
– GustoCo
Dec 6 '18 at 1:45












$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44




$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44












$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44




$begingroup$
To show that $exp(x+y) = exp(x)exp(y)$, consider the function $f(x,y) = exp(x)exp(y)exp(-(x+y))$ and show that both first partials are $0$ everywhere, and $f(0,0)=1$, so the $f(x,y) equiv 1$, and so the equality holds. You can adapt this argument to your situation.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 3:44










1 Answer
1






active

oldest

votes


















0












$begingroup$

Any function whose derivative is 0 for all $x$ is a constant function.



So what you have shown is that $e^{(a+b)x}-e^{ax}e^{bx}$ is a constant function. Any input, $x$, will yield the same output. So for example $e^{5(a+b)}-e^{5a}e^{5b}=e^{89(a+b)}-e^{89a}e^{88b}= e^{-2(a+b)}-e^{-2a}e^{-2b}$. It doesn’t matter what input we choose, it will give us the same thing.



So how do we find the function value? How do we use this information to show the result?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Input the value and get the result.
    $endgroup$
    – William Elliot
    Dec 6 '18 at 10:06






  • 1




    $begingroup$
    Those questions were rhetorical. They were meant for the question asker to think about.
    $endgroup$
    – John B
    Dec 6 '18 at 14:03










  • $begingroup$
    I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
    $endgroup$
    – GustoCo
    Dec 6 '18 at 20:15












  • $begingroup$
    Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
    $endgroup$
    – John B
    Dec 6 '18 at 20:24










  • $begingroup$
    If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
    $endgroup$
    – John B
    Dec 6 '18 at 20:27











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Any function whose derivative is 0 for all $x$ is a constant function.



So what you have shown is that $e^{(a+b)x}-e^{ax}e^{bx}$ is a constant function. Any input, $x$, will yield the same output. So for example $e^{5(a+b)}-e^{5a}e^{5b}=e^{89(a+b)}-e^{89a}e^{88b}= e^{-2(a+b)}-e^{-2a}e^{-2b}$. It doesn’t matter what input we choose, it will give us the same thing.



So how do we find the function value? How do we use this information to show the result?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Input the value and get the result.
    $endgroup$
    – William Elliot
    Dec 6 '18 at 10:06






  • 1




    $begingroup$
    Those questions were rhetorical. They were meant for the question asker to think about.
    $endgroup$
    – John B
    Dec 6 '18 at 14:03










  • $begingroup$
    I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
    $endgroup$
    – GustoCo
    Dec 6 '18 at 20:15












  • $begingroup$
    Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
    $endgroup$
    – John B
    Dec 6 '18 at 20:24










  • $begingroup$
    If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
    $endgroup$
    – John B
    Dec 6 '18 at 20:27
















0












$begingroup$

Any function whose derivative is 0 for all $x$ is a constant function.



So what you have shown is that $e^{(a+b)x}-e^{ax}e^{bx}$ is a constant function. Any input, $x$, will yield the same output. So for example $e^{5(a+b)}-e^{5a}e^{5b}=e^{89(a+b)}-e^{89a}e^{88b}= e^{-2(a+b)}-e^{-2a}e^{-2b}$. It doesn’t matter what input we choose, it will give us the same thing.



So how do we find the function value? How do we use this information to show the result?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Input the value and get the result.
    $endgroup$
    – William Elliot
    Dec 6 '18 at 10:06






  • 1




    $begingroup$
    Those questions were rhetorical. They were meant for the question asker to think about.
    $endgroup$
    – John B
    Dec 6 '18 at 14:03










  • $begingroup$
    I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
    $endgroup$
    – GustoCo
    Dec 6 '18 at 20:15












  • $begingroup$
    Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
    $endgroup$
    – John B
    Dec 6 '18 at 20:24










  • $begingroup$
    If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
    $endgroup$
    – John B
    Dec 6 '18 at 20:27














0












0








0





$begingroup$

Any function whose derivative is 0 for all $x$ is a constant function.



So what you have shown is that $e^{(a+b)x}-e^{ax}e^{bx}$ is a constant function. Any input, $x$, will yield the same output. So for example $e^{5(a+b)}-e^{5a}e^{5b}=e^{89(a+b)}-e^{89a}e^{88b}= e^{-2(a+b)}-e^{-2a}e^{-2b}$. It doesn’t matter what input we choose, it will give us the same thing.



So how do we find the function value? How do we use this information to show the result?






share|cite|improve this answer









$endgroup$



Any function whose derivative is 0 for all $x$ is a constant function.



So what you have shown is that $e^{(a+b)x}-e^{ax}e^{bx}$ is a constant function. Any input, $x$, will yield the same output. So for example $e^{5(a+b)}-e^{5a}e^{5b}=e^{89(a+b)}-e^{89a}e^{88b}= e^{-2(a+b)}-e^{-2a}e^{-2b}$. It doesn’t matter what input we choose, it will give us the same thing.



So how do we find the function value? How do we use this information to show the result?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 5:41









John BJohn B

1766




1766












  • $begingroup$
    Input the value and get the result.
    $endgroup$
    – William Elliot
    Dec 6 '18 at 10:06






  • 1




    $begingroup$
    Those questions were rhetorical. They were meant for the question asker to think about.
    $endgroup$
    – John B
    Dec 6 '18 at 14:03










  • $begingroup$
    I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
    $endgroup$
    – GustoCo
    Dec 6 '18 at 20:15












  • $begingroup$
    Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
    $endgroup$
    – John B
    Dec 6 '18 at 20:24










  • $begingroup$
    If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
    $endgroup$
    – John B
    Dec 6 '18 at 20:27


















  • $begingroup$
    Input the value and get the result.
    $endgroup$
    – William Elliot
    Dec 6 '18 at 10:06






  • 1




    $begingroup$
    Those questions were rhetorical. They were meant for the question asker to think about.
    $endgroup$
    – John B
    Dec 6 '18 at 14:03










  • $begingroup$
    I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
    $endgroup$
    – GustoCo
    Dec 6 '18 at 20:15












  • $begingroup$
    Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
    $endgroup$
    – John B
    Dec 6 '18 at 20:24










  • $begingroup$
    If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
    $endgroup$
    – John B
    Dec 6 '18 at 20:27
















$begingroup$
Input the value and get the result.
$endgroup$
– William Elliot
Dec 6 '18 at 10:06




$begingroup$
Input the value and get the result.
$endgroup$
– William Elliot
Dec 6 '18 at 10:06




1




1




$begingroup$
Those questions were rhetorical. They were meant for the question asker to think about.
$endgroup$
– John B
Dec 6 '18 at 14:03




$begingroup$
Those questions were rhetorical. They were meant for the question asker to think about.
$endgroup$
– John B
Dec 6 '18 at 14:03












$begingroup$
I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
$endgroup$
– GustoCo
Dec 6 '18 at 20:15






$begingroup$
I'm a little confused. I showed exp((a+b)x)−exp(ax)*exp(bx) is 0 when we sub in x = 0. The only reason I subbed in x = 0 is because the definition of the exponential stated at the beginning of the question allows for it. How do I extend this for all value of x? For example, from the definition, I only know that exp((a+b)0) - exp(a*0)*exp(b*0) = 1 - 1 = 0. How can I extend this to exp((a+b)*5) - exp(a*5)*exp(b*5) = 0?
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– GustoCo
Dec 6 '18 at 20:15














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Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
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– John B
Dec 6 '18 at 20:24




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Let me give you examples of constant functions. f(x)=1, g(x)=2, h(x)=-67. If I input 0 into these functions, I get f(0)=1, g(0)=2, h(0)=-67. If I input
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– John B
Dec 6 '18 at 20:24












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If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
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– John B
Dec 6 '18 at 20:27




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If I input 4 into these functions, I get f(4)=1, g(4)=2, h(4)=-67. If I input any number y into these functions, I get f(y)=1, g(y)=2, h(y)=-67. @GustoCo
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– John B
Dec 6 '18 at 20:27


















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