# 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 $g'(F(x))$ $?$

• In your last line of the question what is $h$ ? – Empty Oct 3 '15 at 16:32
• @S.Panja-1729 : Typo Sir,typo. – user118494 Oct 3 '15 at 16:43

Your function is $\psi(x)=F(g(x))-F(a)$ Where $F(x)$ is a primitive of $f(x)$ so the derivative is: $$\psi'(x)=F'(g(x))g'(x)=f(g(x))g'(x)$$
$$\psi (x) = \left( {F \circ g} \right)(x)$$
\eqalign{ & \psi '(x) = F'(g(x))g'(x) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = g'(x)f(g(x)) \cr}
You did the composition wrong! The integral is $F\circ g(x)$ and not $g\circ F(x)$.