Chain rule with a reciprocal trig-of-power function

Let $f(x) = \dfrac{1}{\sin(x^2)}$ . Find $f'(x)$ .

Hint

Rewrite $f(x) = [\sin(x^2)]^{-1}$ so you can use the chain rule with outer function $u^{-1}$ and inner function $u = \sin(x^2)$ . Remember that differentiating $\sin(x^2)$ itself requires another application of the chain rule, since $x^2$ is inside the sine.

Step-by-step solution

Rewrite $f(x) = [\sin(x^2)]^{-1}$ . By the chain rule, the outer function $u^{-1}$ has derivative $-u^{-2}$ , so $f'(x) = -[\sin(x^2)]^{-2} \cdot \dfrac{d}{dx}[\sin(x^2)]$ .

Now differentiate the inner piece $\sin(x^2)$ , again using the chain rule: $\dfrac{d}{dx}[\sin(x^2)] = \cos(x^2)\cdot 2x$ . Substituting back gives $f'(x) = -\dfrac{2x\cos(x^2)}{\sin^2(x^2)}$ .

Chain rule with a reciprocal trig-of-power function

Related problems