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

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!

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 $$dx$$.
$$\frac{\partial f(a,b)}{\partial a} = \frac{\partial}{\partial a}\int_0^1 (ax+b+\frac{1}{1+x^2})^2 dx$$
$$= \int_0^1 \frac{\partial}{\partial a}(ax+b+\frac{1}{1+x^2})^2dx = \int_0^1 2(ax+b+\frac{1}{1+x^2})(ax+b+\frac{1}{1+x^2})' dx = \int_0^1 2(ax+b+\frac{1}{1+x^2})x dx$$
• $2$ is the derivative of $x^2$, and our expression is $g(a)^2$ so the derivative wrt $a$ of this is $2g(a)g'(a)$ – George Dewhirst Apr 22 at 23:44
• How is explicitily $g(a)$ expressed $g(a)=(ax+b+\frac{1}{1+x^2})$ or $g(a)= \int_{0}^{1} (ax+b+\frac{1}{1+x^2})$? – Cos Apr 22 at 23:48
• The first one. We are differentiating $\int_0^1 g(a,x)^2 dx$, we are allowed to take the partial derivative inside of the integral. – George Dewhirst Apr 22 at 23:49
• So $g(a,x)^{2}=(ax+b+\frac{1}{1+x^2})$?? and partial derivative of $f(a,b)$ as I defined is the same of diferentiating $\int_{0}^{1}g(a,x)^{2}$??? And differentiating this last one is $\int_{0}^{1} \frac{\partial{g(a,x)^{2}}}{\partial {a}}$ ???? – Cos Apr 23 at 0:00