To determine
To find:
The derivative of the function using part 1) of the Fundamental Theorem of calculus
Answer
-tan(x)2
Explanation
1) Concept:
i. The Fundamental Theorem of Calculus-Suppose f is continuous on [a, b] then,
if hs=∫auftdt, then h'=fu
ii. Chain rule- Let Fx=fgx, if g is differentiable at x and f is differentiable at g(x) then F'x=f'gx·g'(x)
iii. In Leibnitz notation if y=f(u) and u=g(x) are both differentiable functions then
dydx=dydu·dudx
2) Given:
y=∫xπ4θ tan(θ)dθ
3) Calculation:
Given that
y=∫xπ4θtan(θ)dθ
But
∫xπ4θtan(θ)dθ=-∫π4xθtan(θ)dθ
Substitute it in the above equation, so
y=-∫π4xθtan(θ)dθ
Let u=x and substitute it in the above equation and take derivative of both sides
y'=ddx-∫π4uθtan(θ)dθ
Using chain rule
y'=ddu-∫π4uθtan(θ)dθ·dudx
Use Fundamental Theorem of Calculus part a) with fθ=-θtan(θ) to get
ddu-∫π4uθtan(θ)dθ=-u·tan(u) Thus
y'=-u·tan(u)·dudx
But u=x
Differentiate u
dudx=12x
Substitute u=x and dudx=12x in y'. Thus
y'=-x·tan(x)·12x
Simplify
y'=-tan(x)2
Therefore,y'=-tan(x)2
Conclusion:
Therefore,y'=-tan(x)2