Given information: The function is tanx+tanx·sec2x

Let’s remind of the fundamental theorem of calculus part 1:

The fundamental theorem of calculus part 1: If f is continuous on [a,b] then the function of g defined by

  g(x)=∫axf(t) dt where a≤x≤b is continuous on [a,b] and differentiable on (a,b) and g′(x)=f(x) .

We have, y=∫0tan xt+t dt

Since the upper limit of integration is not y , we apply the chain rule

Let , y=f(u) , u=tanx

Then, u=tanx , u′=sec2x

Consider the new function

  f(u)=∫0ut+t dt

Differentiation with respect to u,

  ⇒dduf(u)=ddu[∫0ut+t dt]

  ⇒dduf(u)

  =u+u

  ⇒f′(u)=u+u

Now y=f(u)

  ⇒y′=[f(tanx)u]′

  ⇒y′=f′(tanx)·u′

  y′=tanx+tanx·sec2x

Hence, the derivative is tanx+tanx·sec2x .