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:


But this is incorrect.

Any help?

  • 1
    $\begingroup$ Tip: Try writing \sin (x) instead of sin (x) inside the dollar signs. It produces $\sin (x)$ in comparison to $sin (x)$. $\endgroup$ – For the love of maths Jan 20 '18 at 6:07

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

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


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

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,


  • $\begingroup$ This is the equation for which I calculated the derivative. $cos\left(sin\left(x\right)^4\right)+sin\left(x\right)$ Is my calculation incorrect? Am I supposed to calculate the derivative of a different equation? Thank you for your help! $\endgroup$ – sktsasus Jan 20 '18 at 6:16
  • $\begingroup$ Are you able to identify $f$ and $g$ in $h(x)=f(g(x))$ in this particular case? $\endgroup$ – Siong Thye Goh Jan 20 '18 at 6:24
  • $\begingroup$ Yes, I believe so. Would $f$ be $cos$ and $g$ be $sin$? I think I am struggling to figure out which equation I am supposed to derive. Your answer (which is correct) says that I only find the derivative of the very last part which is $sin(x)$. Why do we ignore the first part of the equation? $\endgroup$ – sktsasus Jan 20 '18 at 6:27
  • $\begingroup$ actually $f(x)= \int_{-4}^x (\cos(t^4)+t) \, dt$ and $g(x)=\sin x$. We use fundamental theorem in differentiating $f$ then substitute $g(x)$ in there using chain rule and multiply by the derivative of $g$. $\endgroup$ – Siong Thye Goh Jan 20 '18 at 6:29
  • $\begingroup$ OK I understand. Thank you for your help! $\endgroup$ – sktsasus Jan 20 '18 at 6:32

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