How do you calculate the derivative of #(cos(t^2)+t) dt# from #t=-5# to #t=sinx#?

1 Answer
Jun 26, 2016

Answer:

#= cos x( cos (sin x)^2 + sin x)#

Explanation:

i think you mean this

#d/dx int_{-5}^{sin x} \ cos t^2 + t\ dt#

We can use the Leibnitz Rule for differentiating under the integral sign

Wikipedia

which simplifies considerably here to

#d/dx int_{-5}^{sin x} \ cos t^2 + t\ dt#

#= cos (sin x)^2 (sin x)' + sin x (sin x)'#

#= cos (sin x)^2 cos x + sin x cos x#

#= cos x( cos (sin x)^2 + sin x)#

if that seems overwhelming, we can look at the Fundamental Theorem of Calculus, specifically

if #F(x) = int_a^x f(t) \ dt#, then #F'(x) = f(x)#

this can be extended using the chain rule to

if #F(x) = int_a^{v(x)} f(t) \ dt#, then #F'(x) = f(v(x)) * v'(x)#

that's all we did here