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[(t^5)/(5)\right]^(15)_x \\ compute\text{ the boundaries} \\ f(x)=151875-(x^5)/(5) \end{gathered}`

Step 3

Find f'(x)

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