2.4 · Implicit and inverse-function derivatives

Differentiate both sides with respect to x

If an equation defines $\displaystyle y$ as a function of $\displaystyle x$, differentiating $\displaystyle y^2$ gives $\displaystyle 2y\frac{dy}{dx}$. A $\displaystyle y$ term changes with $\displaystyle x$, so its derivative cannot be treated as a constant.

For $\displaystyle x^2+y^2=25$, differentiation gives $\displaystyle 2x+2yy'=0$. Solving yields $\displaystyle y'=-x/y$ where $\displaystyle y\ne0$. Check that a requested point lies on the relation before finding a tangent.

$$\frac{d}{dx}y^n=ny^{n-1}\frac{dy}{dx}$$

Differentiate an inverse relationship

If $\displaystyle g=f^{-1}$, then $\displaystyle f(g(x))=x$. The chain rule gives $\displaystyle f'(g(x))g'(x)=1$, so the inverse derivative is the reciprocal of the original derivative at the corresponding input, provided it is nonzero.

If $\displaystyle f(2)=5$ and $\displaystyle f'(2)=3$, then $\displaystyle g(5)=2$ and $\displaystyle g'(5)=1/3$. Do not evaluate $\displaystyle f'$ at $\displaystyle 5$ just because the inverse input is $\displaystyle 5$.

$$(f^{-1})'(b)=\frac{1}{f'(a)},\qquad f(a)=b$$

Use inverse-trigonometric rules carefully

The derivatives are $\displaystyle (\arcsin x)'=1/\sqrt{1-x^2}$ for $\displaystyle |x|<1$, $\displaystyle (\arccos x)'=-1/\sqrt{1-x^2}$ for $\displaystyle |x|<1$, and $\displaystyle (\arctan x)'=1/(1+x^2)$ for real $\displaystyle x$. Composed inputs also need the chain rule.

An inverse trigonometric function is not a reciprocal: $\displaystyle \arcsin x$ returns an angle, whereas $\displaystyle \csc x$ is $\displaystyle 1/\sin x$. After implicit differentiation, higher derivatives require another application of the chain rule and substitution for $\displaystyle y'$ when appropriate.