How would I find the following derivative?

Question

Solve for $h'(x)$ using the fundamental theorem of calculus. $$h(x) = \int _{-4}^{\sin\left(x\right)}\left(\cos\left(t^4\right)+t\right)\,dt$$

I tried to do this by plugging in the $\sin(x)$ into both of the $t$'s and then tried to calculate the derivative of that.

This is the derivative I calculated:

$$-4\sin^3(x)\sin\left(\sin^4(x)\right)\cos(x)+\cos(x)$$

But this is incorrect.

Any help?

Answer 1

$$h(x) = \int_{-4}^{\sin(x)} (\cos(t^4)+t) \, dt$$

Applying chain rule,

you should get $$h'(x)=\left(\cos (\color{blue}{\sin}^4\color{blue}{(x)})+\color{blue}{\sin(x)}\right) \color{green}{\frac{d}{dx}\sin (x)}$$

Recall that in chain rule,

$$\frac{d}{dx}f(g(x))=f'(\color{blue}{g(x)})\color{green}{g'(x)}$$

Answer 2

The second fundamental theorem of calculus states $$\frac{d}{dx}\int_c^{u(x)} f(t)\,dt=f(u)\cdot\frac{du}{dx}$$

Applying this to your example gives

$$h’(x)= \left[\cos\left(\sin^4 x\right)+\sin x\right] \frac{d}{dx}(\sin x) = \left[\cos\left(\sin^4 x\right)+\sin x\right]\cos x$$