# How to solve such question.

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$$\frac{d}{dx}\int_{1}^{x^4} sect\space dt$$
in Calculus
0
integration of sect is ln|sect + tant| thereafter substitute t2=x^4 and t1=1

Then you got result in terms of x which you have to do differentiation wrt x
3

it is based on lebnitz's rule:

### $\frac{\mathrm{d} }{\mathrm{d} x}\int_{v(x))}^{u(x))}f(t)dt$=f(u(x))$\frac{\mathrm{d} }{\mathrm{d} x}$(u(x))-f(v(x))$\frac{\mathrm{d} }{\mathrm{d} x}$(v(x))

=sec($x^{4}$)*4$x^{3}$

0

@newdreamz a1-z0 How you got sec in the answer.

I applied the same formula and got $x^44x^3$

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