Question

f(x) = integrate t ^ 4 dt from x to 15 * then; f^ prime (x)= Box

Answer 1

Step 1

Given;

Step 2

$\begin{gathered} f(x)=\int_x^{15}t^4dt \\ Apply\text{ the power rule } \\ =\left[\frac{t^5}{5}\right]^{15}_x \\ compute\text{ the boundaries} \\ f(x)=151875-\frac{x^5}{5} \end{gathered}$

Step 3

Find f'(x)

$\begin{gathered} \frac{d}{dx}\left(151875-\frac{x^5}{5}\right) \\ \mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g' \\ =\frac{d}{dx}\left(151875\right)-\frac{d}{dx}\left(\frac{x^5}{5}\right) \\ =\frac{d}{dx}\left(151875\right)=0 \\ =\frac{d}{dx}\left(\frac{x^5}{5}\right)=x^4 \end{gathered}$

Thus;

$\begin{gathered} f^{^{\prime}}(x)=0-x^4 \\ f^{^{\prime}}(x)=-x^4 \end{gathered}$

Answer;

$f^{\prime}(x)=-x^4$