Question

Describe how one can find the derivative of the function given below, using Part 1 of Fundamental Theorem of Calculus?
$ y = \int {{{(6 + {v^8})}^3}dv} $ ; from $ 1 $ to $ \cos (x) $

Answer 1

Hint: We can see that we are given an integral with a higher degree, so rather than directly solving the integral, we must see if we are able to use any theorem to get to the solution. The steps you can use would be, first state the theorem, relate the question to the theorem, look at the limits and if they are complex then go ahead with the substitution method, and complete it using chain rule.

Complete step by step solution:
Let us start off by writing down the question, so that we can understand what our goal is;
We need to find derivative of: $ y = \int\limits_1^{\cos (x)} {{{(6 + {v^8})}^3}dv} $
Now we know that the Fundamental Theorem of Calculus writes;
For any given function defined in the interval $ [a,b] $ ;
 $ \dfrac{d}{{dx}}(\int\limits_c^x {f(v)dv)} = f(x);\; $ $ \forall \;x \in [a,b] $
Let us separately write the antiderivative of the function $ f $ in the interval $ [a,b] $ , let it be $ S(x) $ ;
 $ S(x) = \int\limits_c^x {f(v)dv} $
Now in our question, since a trigonometric function is one of the limits we must proceed to substitution so that it makes solving the integral easier;
Let the functions be defined as: $ $
 $ S(x) = \int\limits_1^x {{{(6 + {v^8})}^3}} $
So now we can put another function to represent the inner term;
  $ h(x) = \cos (x) $
Now our goal has changed to find derivative of a function within a function, written as;
 $ \dfrac{d}{{dx}}(S(h(x)) $
By the formula of chain rule, we can write;
 $ \dfrac{d}{{dx}}(S(h(x)) = S'(h(x)) \times h'(x) $
So now we solve to find the terms on the right hand side;
 $ S'(x) = {(6 + {x^8})^3} $
 $ h'(x) = - \sin (x) $
Substituting the terms on the right hand side will be;
 $ \dfrac{d}{{dx}}(S(h(x)) = - {(6 + {\cos ^8}(x))^3} \times \sin (x) $
 $ \therefore $ The final answer for this question will be;
 $ \dfrac{d}{{dx}}(\int\limits_1^{\cos (x)} {{{(6 + {v^8})}^3}dv} ) = - {(6 + {\cos ^8}(x))^3} \times \sin (x) $
So, the correct answer is “ $ \dfrac{d}{{dx}}(\int\limits_1^{\cos (x)} {{{(6 + {v^8})}^3}dv} ) = - {(6 + {\cos ^8}(x))^3} \times \sin (x) $ ”.

Note: Solving problems is made easier using this theorem, The Fundamental Theorem of Calculus. Calculations become simpler, since chain rules and substitution is used to solve problems which have complicated limits, like trigonometric functions. We can use any of the two parts of this theorem, depending on whether we wish to solve the derivative of the antiderivative or to solve the integral between constant limits.