Derivatives of Implicit Functions

Method. (1) Differentiate every term of the equation with respect to $x$, treating $y$ throughout as a differentiable function of $x$ — so any term containing $y$ needs the chain rule (an extra factor of $\dfrac{dy}{dx}$ appears each time $y$ is differentiated), and any term where $x$ and $y$ are multiplied together (like $xy^3$) additionally needs the product rule. (2) Collect every term containing $\dfrac{dy}{dx}$ onto one side of the resulting equation, factor $\dfrac{dy}{dx}$ out, and solve for it. The result is generally a formula in both $x$ and $y$ — which is normal and expected for an implicit relation.

Worked Example 1 — find $\dfrac{dy}{dx}$:

1. $x^5+xy^3+x^2y+y^4=4$: differentiating term by term, $5x^4+x(3y^2)\dfrac{dy}{dx}+y^3+x^2\dfrac{dy}{dx}+2xy+4y^3\dfrac{dy}{dx}=0$. Collecting the $\dfrac{dy}{dx}$ terms: $(x^2+3xy^2+4y^3)\dfrac{dy}{dx}=-(5x^4+2xy+y^3)$, so $\dfrac{dy}{dx}=-\dfrac{5x^4+2xy+y^3}{x^2+3xy^2+4y^3}.$
2. $y^3+\cos(xy)=x^2-\sin(x+y)$: differentiating, $3y^2\dfrac{dy}{dx}-\sin(xy)\!\left(x\dfrac{dy}{dx}+y\right)=2x-\cos(x+y)\!\left(1+\dfrac{dy}{dx}\right)$. Collecting the $\dfrac{dy}{dx}$ terms: $[3y^2-x\sin(xy)+\cos(x+y)]\dfrac{dy}{dx}=2x+y\sin(xy)-\cos(x+y)$, so $\dfrac{dy}{dx}=\dfrac{2x+y\sin(xy)-\cos(x+y)}{3y^2-x\sin(xy)+\cos(x+y)}.$
3. $x^2+e^{xy}=y^2+\log(x+y)$: differentiating, $2x+e^{xy}\!\left(x\dfrac{dy}{dx}+y\right)=2y\dfrac{dy}{dx}+\dfrac1{x+y}\!\left(1+\dfrac{dy}{dx}\right)$. Collecting the $\dfrac{dy}{dx}$ terms and clearing the $\frac1{x+y}$ factor by multiplying through by $(x+y)$: $\dfrac{dy}{dx}=\dfrac{2x(x+y)+ye^{xy}(x+y)-1}{2y(x+y)-xe^{xy}(x+y)+1}.$ Worked Example 2. If $x^my^n=(x+y)^{m+n}$, show $\dfrac{dy}{dx}=\dfrac yx$. Taking logs of both sides: $m\log x+n\log y=(m+n)\log(x+y)$. Differentiating implicitly: $\dfrac mx+\dfrac ny\dfrac{dy}{dx}=\dfrac{m+n}{x+y}\!\left(1+\dfrac{dy}{dx}\right)$. Collecting the $\dfrac{dy}{dx}$ terms and simplifying the resulting fraction (using $mx+ny$ cancellations after cross-multiplying) leads directly to $\dfrac{dy}{dx}=\dfrac yx$. This 'take logs first' trick — turning a product-equals-a-power equation into a linear relation between $\log x$ and $\log y$ — is exactly what makes the whole family of Exercise 1.3(4) identities collapse so cleanly to $y/x$. Worked Example 3. If $\sin\!\left(\dfrac{px^m-qy^m}{px^m+qy^m}\right)=r$ (a constant), show $\dfrac{dy}{dx}=\dfrac yx$. Writing $t=\sin^{-1}r$ (a constant) turns the equation into $\dfrac{px^m-qy^m}{px^m+qy^m}=t$, which rearranges (cross-multiplying and collecting) to $y^m=s\cdot x^m$ for a constant $s=\dfrac{p(1-t)}{q(1+t)}$. Differentiating $y^m=sx^m$ implicitly: $my^{m-1}\dfrac{dy}{dx}=smx^{m-1}$, so $\dfrac{dy}{dx}=s\cdot\dfrac{x^{m-1}}{y^{m-1}}=\dfrac{y^m}{x^m}\cdot\dfrac{x^{m-1}}{y^{m-1}}=\dfrac yx$. Worked Example 4. If $\sec^{-1}\!\left(\dfrac{x^3+y^3}{x^3-y^3}\right)=2a$ (a constant, not eliminated since the target answer contains $a$), show $\dfrac{dy}{dx}=\dfrac{x^2\tan^2a}{y^2}$. Rewriting as $\cos^{-1}\!\left(\dfrac{x^3-y^3}{x^3+y^3}\right)=2a$, so $\dfrac{x^3-y^3}{x^3+y^3}=\cos2a$. Cross-multiplying and collecting the $x^3,y^3$ terms: $x^3(1-\cos2a)=y^3(1+\cos2a)$, i.e. $y^3=\dfrac{2\sin^2a}{2\cos^2a}x^3=(\tan^2a)x^3$. Differentiating: $3y^2\dfrac{dy}{dx}=3(\tan^2a)x^2$, so $\dfrac{dy}{dx}=\dfrac{x^2\tan^2a}{y^2}$. …