# How do you find the derivative of G(x)=int (tan(t^2))dt from [1,x]?

Feb 21, 2017

$\frac{d}{\mathrm{dx}} G \left(x\right) = G ' \left(x\right) = \tan \left({x}^{2}\right)$

#### Explanation:

If asked to find the derivative of an integral then you should not evaluate the integral, instead use the the fundamental theorem of Calculus, which formally states that:

$\frac{d}{\mathrm{dx}} \setminus {\int}_{a}^{x} \setminus f \left(\tau\right) \setminus d \tau = f \left(x\right)$

(ie the derivative of an integral gives us the original function back, or that differentiation undoes the result of integration.). Note that the lower limit can be any constant, and that the upper limit is $x$ which is the same as the variable is the resulting solution $f \left(x\right)$ and that $f \left(x\right)$ is independent of the variable of integration $\tau$.

If the upper limit is itself function of $x$ (eg $g \left(x\right)$) then we must apply the chain rule to get a valid solution, as

$\frac{d}{\mathrm{dx}} \setminus {\int}_{a}^{g \left(x\right)} f \left(\tau\right) \setminus d \tau = \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} \frac{d}{\mathrm{dg} \left(x\right)} {\int}_{a}^{g \left(x\right)} f \left(\tau\right) \setminus d \tau$
$\text{ } = g ' \left(x\right) f \left(g \left(x\right)\right)$

And so we have;

$\setminus \setminus \setminus \setminus \setminus \setminus G \left(x\right) = {\int}_{1}^{x} \setminus \tan \left({t}^{2}\right) \setminus \mathrm{dt}$

$\therefore G ' \left(x\right) = \frac{d}{\mathrm{dx}} \setminus {\int}_{1}^{x} \setminus \tan \left({t}^{2}\right) \setminus \mathrm{dt}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \tan \left({x}^{2}\right)$