Find the derivative of $\int_a^{g(x)} f(t)dt $ wrt $x$ Question :
Let $$f:[a,b]\rightarrow \mathbb R$$ be continuous  and $$g:[c,d]\rightarrow \mathbb R$$ be  differentiable .  Define  $$\psi(x) := \int_a^{g(x)} f(t)dt $$ . Prove  that $\psi$ is  differentiable and  compute  the  derivative .
My Attempt :
Define $$F(x)=\int_a^x f(t)dt$$
Then by Second Fundamental Theorem   of  Calculus , we  have $F$  is  differentiable  and  $F'=f$ at  the points  of  continuity  of  $f$ .
Then $$\psi(x)=(g\circ F)(x)$$
Both  being  differentiable ,  their  composition  is  so   and  hence  $\psi$  is  differentiable. And  by  Chain Rule  we  have , $$\psi'(x)=g'(F(x))F'(x) \\ =(g'(F(x)))\cdot f(x)$$
Upto  this  is  I  think  ok. But  now  how to  simplify $g'(F(x))$ $?$
Your function is  $\psi(x)=F(g(x))-F(a)$ Where $F(x)$ is a primitive of $f(x)$ so the derivative is:
$$
\psi'(x)=F'(g(x))g'(x)=f(g(x))g'(x)
$$
You did the composition wrong! The integral is $F\circ g(x)$ and not $g\circ F(x)$.