Calculate the derivative of $f(x)=\int_x^{x^2}\tan(x+y)\,dy$ on the open interval $(0,\frac{\pi}{4})$.

Question

This appears to be a chain rule question, but I'm struggling with the setup. My attempt is as follows:

Let $F:\mathbb{R}^2\rightarrow\mathbb{R}$,

$F(u,v)=\int_u^{u^2}$tan$(v+y)dy$,

$u:\mathbb{R}\rightarrow\mathbb{R}^2$,

and $u(x)=(u(x),v(x))=(x,x)$.

Then writing $f(x)=(F\circ u)(x)$ allows for use of the chain rule:

$D(F\circ u)(x)=DF(u(x),v(x))Du(x)=\left[\frac{\partial F}{\partial u}\mbox{ }\frac{\partial F}{\partial v}\right]\begin{bmatrix}u'(x)\\v'(x)\end{bmatrix}=\frac{\partial F}{\partial u}(u(x))u'(x)+\frac{\partial F}{\partial v}(u(x))v'(x)$

Then by the Fundamental Theorem of Calculus,

$\frac{\partial F}{\partial u}=\frac{\partial}{\partial u}\int_u^{u^2}tan(v+y)dy=\mbox{tan}(v+u)=\mbox{tan}(x+x)=\mbox{tan(2x)}$

And similarly,

$\frac{\partial F}{\partial v}=\frac{\partial}{\partial v}\int_u^{u^2}\mbox{tan}(v+y)dy=\int_u^{u^2}\frac{\partial}{\partial v}\mbox{tan}(v+y)dy=\int_x^{x^2}\frac{\partial}{\partial x}\mbox{tan}(x+y)dy=\int_x^{x^2}\mbox{sec}^2(x+y)dy$

However at this point, I'm relatively convinced I'm making a major mistake, either with my entire substitution or with the way I'm handling the boundaries. Any advice is much appreciated.

Answer 1

Using Leibniz integral rule:

\begin{align} f'(x) &= 2x\tan(x+x^2) - \tan(2x) + \int_x^{x^2} \frac{dy}{\cos^2(x+y)}\\ &= 2x\tan(x+x^2) - \tan(2x)+ \tan(x+y)\Big|_x^{x^2}\\ &= (2x+1)\tan(x+x^2) - 2\tan(2x) \end{align}

Answer 2

Based upon your correction, I have revised my answer. $f(x)=\int_x^{x^2}tan(x+y)dy=ln(cos(x+x^2))-ln(cos(2x))$. Therefore $f'(x)=(2x+1)tan(x+x^2)-2tan(2x)$.

Answer 3

Let me try: \begin{align} f'(x) &=\partial_x \int\limits_x^{x^2}\tan(x+y)\,dy \\ &= \tan(x+ x^2) (x^2)' - \tan(x + x)(x)' + \int\limits_x^{x^2}\partial_x \tan(x+y)\,dy \\ &= 2x \tan(x+ x^2) - \tan(2x) + \int\limits_x^{x^2}(1+ \tan^2(x+y)) dy \\ &= 2x \tan(x+ x^2) - \tan(2x) + \left[\tan(x + y) \right]_{y=x}^{y=x^2} \\ &= 2x \tan(x+x^2) - \tan(2x) + \tan(x+x^2) - \tan(2x) \\ &= (2x+1) \tan(x+x^2) - 2\tan(2x) \end{align}

Answer 4

Hint:

Let $$F(x,t)= \int_{0}^x \tan(t+y)dy$$

Then we have $$f(x)= F(x,x^2)-F(x,x)$$

NOTE THAT $\partial_x F(x,t) = \tan(t+x)$ and $\partial_t F(x,t) = \tan(x+t)- \tan(t)$

can you conclude?