Computing partial derivatives of $f(a,b)= \int_{0}^{1}(ax+b+\frac{1}{1+x^2})^{2}dx$ using chain rule.

Question

Let $f(a,b)= \int_{0}^{1}(ax+b+\frac{1}{1+x^2})^{2}dx$. I want to compute $\frac{\partial{f(a,b)}}{\partial{a}}$ and $\frac{\partial{f}(a,b)}{\partial{b}}$. I was told in the text that $$\frac{\partial{f}(a,b)}{\partial{a}}=2. \int_{0}^{1}(ax+b+\frac{1}{1+x^{2}})dx x$$ and $$\frac{\partial{f}(a,b)}{\partial{b}}=2. \int_{0}^{1}(ax+b+\frac{1}{1+x^{2}})dx$$ because the Chain rule but I cannot justify this. I was thinking this holds because if $f(a,b)=F(H(a))$ where $$F(t)=\int_{0}^{1}(t+b+\frac{1}{1+x^2})^{2}dx$$ and $H(a)=ax$, then $\frac{\partial{f(a,b)}}{\partial{a}}=F'(H(a))H'(a)$ but there are counterexamples where $f(a,b) \neq F(H(a)$. Can anyone help me fill the gaps about how these partial derivatives were obtained?? Thanks!

Answer 1

It is because the derivative with respect to $a$ of $(ax+b+\frac{1}{1+x^2}$) is $x$. Best keep the $x$ inside the integrand before the $dx$.

$\frac{\partial f(a,b)}{\partial a} = \frac{\partial}{\partial a}\int_0^1 (ax+b+\frac{1}{1+x^2})^2 dx $

$= \int_0^1 \frac{\partial}{\partial a}(ax+b+\frac{1}{1+x^2})^2dx = \int_0^1 2(ax+b+\frac{1}{1+x^2})(ax+b+\frac{1}{1+x^2})' dx = \int_0^1 2(ax+b+\frac{1}{1+x^2})x dx $