Use part one of the fundamental theorem of calculus to find the derivative of the function.
$g (x) = \int_{0}^{x} \sqrt{t^{3} + t^{5}} d t$
$g^{'} (x) =$

Answer

Expert–verified

Hide Steps

Answer

$g^{'} (x) = \sqrt{x^{5} + x^{3}}$ is the final answer

Steps

Step 1 ：Given the function $g (x) = \int_{0}^{x} \sqrt{t^{3} + t^{5}} d t$

Step 2 ：We are asked to find the derivative of this function $g (x)$

Step 3 ：According to the Fundamental Theorem of Calculus, the derivative of this function $g (x)$ is just the function inside the integral evaluated at x

Step 4 ：So, we need to replace t with x in the function inside the integral to get the derivative of $g (x)$

Step 5 ：Substituting $t = x$ in the function inside the integral, we get $g^{'} (x) = \sqrt{x^{5} + x^{3}}$

Step 6 ： $g^{'} (x) = \sqrt{x^{5} + x^{3}}$ is the final answer

Use part one of the fundamental theorem of calculus to find the derivative of the function.g(x)=∫0xt3+t5dtg′(x)=

Answer

Popular Problems

Use part one of the fundamental theorem of calculus to find the derivative of the function.
$g (x) = \int_{0}^{x} \sqrt{t^{3} + t^{5}} d t$
$g^{'} (x) =$