implicit differentiation yielding different expressions
$begingroup$
I was studying Thomas's Calculus book and attempted a question using implicit differentiation.
$$x^3=frac{2x-y}{x+3y}$$
I differentiated both sides directly, using the quotient rule on the RHS to obtain. $$ frac{dy}{dx}=frac{7y-3x^2(x+3y)^2}{7x} hspace{1cm}(1)$$
The steps in the solution were to bring $x+3y$ to the LHS and differentiate every term to get $$ frac{dy}{dx}=frac{2-4x^3-9x^2y}{3x^3+1}hspace{1cm} (2)$$
I was confused at first, and guessed that the two must be equal. And could show it using these steps. Expanding the numerator of $(1)$ gives $$frac{dy}{dx}=frac{7y-3x^4-18x^3y-27x^2y^2}{7x}hspace{1cm} (1)$$
Replacing $x^3$ in the denominator of $(2)$ by $frac{2x-y}{x+3y}$ and some simplification gives$$ begin{align}frac{dy}{dx}&=frac{2x+6y-4x^4-21x^3y-27x^2y^2}{7x}\&=frac{7y-3x^4-18x^3-27x^2y^2+(2x-y-x^4-3x^3y)}{7x}end{align}hspace{1cm} (2)$$
The sum in the brackets is zero by $$2x-y-x^4-3x^3y=2x-y-x^3(x+3y)=2x-y-(2x-y)=0$$
Therefore $(1)=(2)$. Firstly, I am still not entirely sure $textbf{why}$ we get two expressions which look completely different in the first place. Moreover is there any $textbf{better way}$ to show that they are equal besides the fortuitous, trial and error, method I used? Many many thanks.
implicit-differentiation
asked Jan 2 at 11:49
mattmatt
392213
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add a comment |
$begingroup$
I was studying Thomas's Calculus book and attempted a question using implicit differentiation.
$$x^3=frac{2x-y}{x+3y}$$
I differentiated both sides directly, using the quotient rule on the RHS to obtain. $$ frac{dy}{dx}=frac{7y-3x^2(x+3y)^2}{7x} hspace{1cm}(1)$$
The steps in the solution were to bring $x+3y$ to the LHS and differentiate every term to get $$ frac{dy}{dx}=frac{2-4x^3-9x^2y}{3x^3+1}hspace{1cm} (2)$$
I was confused at first, and guessed that the two must be equal. And could show it using these steps. Expanding the numerator of $(1)$ gives $$frac{dy}{dx}=frac{7y-3x^4-18x^3y-27x^2y^2}{7x}hspace{1cm} (1)$$
Replacing $x^3$ in the denominator of $(2)$ by $frac{2x-y}{x+3y}$ and some simplification gives$$ begin{align}frac{dy}{dx}&=frac{2x+6y-4x^4-21x^3y-27x^2y^2}{7x}\&=frac{7y-3x^4-18x^3-27x^2y^2+(2x-y-x^4-3x^3y)}{7x}end{align}hspace{1cm} (2)$$
The sum in the brackets is zero by $$2x-y-x^4-3x^3y=2x-y-x^3(x+3y)=2x-y-(2x-y)=0$$
Therefore $(1)=(2)$. Firstly, I am still not entirely sure $textbf{why}$ we get two expressions which look completely different in the first place. Moreover is there any $textbf{better way}$ to show that they are equal besides the fortuitous, trial and error, method I used? Many many thanks.
implicit-differentiation
asked Jan 2 at 11:49
mattmatt
392213
$endgroup$
add a comment |
$begingroup$
I was studying Thomas's Calculus book and attempted a question using implicit differentiation.
$$x^3=frac{2x-y}{x+3y}$$
I differentiated both sides directly, using the quotient rule on the RHS to obtain. $$ frac{dy}{dx}=frac{7y-3x^2(x+3y)^2}{7x} hspace{1cm}(1)$$
The steps in the solution were to bring $x+3y$ to the LHS and differentiate every term to get $$ frac{dy}{dx}=frac{2-4x^3-9x^2y}{3x^3+1}hspace{1cm} (2)$$
I was confused at first, and guessed that the two must be equal. And could show it using these steps. Expanding the numerator of $(1)$ gives $$frac{dy}{dx}=frac{7y-3x^4-18x^3y-27x^2y^2}{7x}hspace{1cm} (1)$$
Replacing $x^3$ in the denominator of $(2)$ by $frac{2x-y}{x+3y}$ and some simplification gives$$ begin{align}frac{dy}{dx}&=frac{2x+6y-4x^4-21x^3y-27x^2y^2}{7x}\&=frac{7y-3x^4-18x^3-27x^2y^2+(2x-y-x^4-3x^3y)}{7x}end{align}hspace{1cm} (2)$$
The sum in the brackets is zero by $$2x-y-x^4-3x^3y=2x-y-x^3(x+3y)=2x-y-(2x-y)=0$$
Therefore $(1)=(2)$. Firstly, I am still not entirely sure $textbf{why}$ we get two expressions which look completely different in the first place. Moreover is there any $textbf{better way}$ to show that they are equal besides the fortuitous, trial and error, method I used? Many many thanks.
implicit-differentiation
asked Jan 2 at 11:49
mattmatt
392213
$endgroup$
I was studying Thomas's Calculus book and attempted a question using implicit differentiation.
$$x^3=frac{2x-y}{x+3y}$$
I differentiated both sides directly, using the quotient rule on the RHS to obtain. $$ frac{dy}{dx}=frac{7y-3x^2(x+3y)^2}{7x} hspace{1cm}(1)$$
The steps in the solution were to bring $x+3y$ to the LHS and differentiate every term to get $$ frac{dy}{dx}=frac{2-4x^3-9x^2y}{3x^3+1}hspace{1cm} (2)$$
I was confused at first, and guessed that the two must be equal. And could show it using these steps. Expanding the numerator of $(1)$ gives $$frac{dy}{dx}=frac{7y-3x^4-18x^3y-27x^2y^2}{7x}hspace{1cm} (1)$$
Replacing $x^3$ in the denominator of $(2)$ by $frac{2x-y}{x+3y}$ and some simplification gives$$ begin{align}frac{dy}{dx}&=frac{2x+6y-4x^4-21x^3y-27x^2y^2}{7x}\&=frac{7y-3x^4-18x^3-27x^2y^2+(2x-y-x^4-3x^3y)}{7x}end{align}hspace{1cm} (2)$$
The sum in the brackets is zero by $$2x-y-x^4-3x^3y=2x-y-x^3(x+3y)=2x-y-(2x-y)=0$$
Therefore $(1)=(2)$. Firstly, I am still not entirely sure $textbf{why}$ we get two expressions which look completely different in the first place. Moreover is there any $textbf{better way}$ to show that they are equal besides the fortuitous, trial and error, method I used? Many many thanks.
implicit-differentiation
implicit-differentiation
asked Jan 2 at 11:49
mattmatt
392213
asked Jan 2 at 11:49
mattmatt
392213
asked Jan 2 at 11:49
mattmatt
392213
asked Jan 2 at 11:49
mattmatt
392213
asked Jan 2 at 11:49
mattmatt
392213
392213
add a comment |
add a comment |
1 Answer
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Solving your equation for $y$ we get
$$y=frac{2x-x^4}{3x^3+1}$$ now you can use the quotient rule.
answered Jan 2 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
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Thanks, I did not notice that, I was wondering about the case when we can not solve for y.
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– matt
Jan 2 at 12:24
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You will need the condition $$3x^3+1neq 0$$
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– Dr. Sonnhard Graubner
Jan 2 at 12:31
add a comment |
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$begingroup$
Solving your equation for $y$ we get
$$y=frac{2x-x^4}{3x^3+1}$$ now you can use the quotient rule.
answered Jan 2 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
$begingroup$
Thanks, I did not notice that, I was wondering about the case when we can not solve for y.
$endgroup$
– matt
Jan 2 at 12:24
$begingroup$
You will need the condition $$3x^3+1neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 12:31
add a comment |
$begingroup$
Solving your equation for $y$ we get
$$y=frac{2x-x^4}{3x^3+1}$$ now you can use the quotient rule.
answered Jan 2 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
$begingroup$
Thanks, I did not notice that, I was wondering about the case when we can not solve for y.
$endgroup$
– matt
Jan 2 at 12:24
$begingroup$
You will need the condition $$3x^3+1neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 12:31
add a comment |
$begingroup$
Solving your equation for $y$ we get
$$y=frac{2x-x^4}{3x^3+1}$$ now you can use the quotient rule.
answered Jan 2 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
Solving your equation for $y$ we get
$$y=frac{2x-x^4}{3x^3+1}$$ now you can use the quotient rule.
answered Jan 2 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
answered Jan 2 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
answered Jan 2 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
answered Jan 2 at 11:59
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
76.7k42866
$begingroup$
Thanks, I did not notice that, I was wondering about the case when we can not solve for y.
$endgroup$
– matt
Jan 2 at 12:24
$begingroup$
You will need the condition $$3x^3+1neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 12:31
add a comment |
$begingroup$
Thanks, I did not notice that, I was wondering about the case when we can not solve for y.
$endgroup$
– matt
Jan 2 at 12:24
$begingroup$
You will need the condition $$3x^3+1neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 12:31
$begingroup$
Thanks, I did not notice that, I was wondering about the case when we can not solve for y.
$endgroup$
– matt
Jan 2 at 12:24
$begingroup$
Thanks, I did not notice that, I was wondering about the case when we can not solve for y.
$endgroup$
– matt
Jan 2 at 12:24
$begingroup$
You will need the condition $$3x^3+1neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 12:31
$begingroup$
You will need the condition $$3x^3+1neq 0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 12:31
add a comment |
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