Copyright

1. What is the derivative of cos x / x^2? 2.Integrate: integral {6 x^5} / {(8 + x^6)^5} dx.

Question:

1. What is the derivative of {eq}\displaystyle \dfrac {\cos x}{x^2} {/eq}?

2.Integrate: {eq}\displaystyle \int \dfrac {6 x^5} {(8 + x^6)^5}\ dx {/eq}.

Differentiation and Integration:

For finding the derivative of {eq}\displaystyle \frac{f(x)}{g(x)} {/eq}, use the quotient rule of the derivative. It is given by {eq}\displaystyle {\left( {\frac{f}{g}} \right)^\prime } = \frac{{f'\,g - f\,g'}}{{{g^2}}} {/eq}

Integration is the process of finding the function {eq}f(x) {/eq} given the function {eq}f'(x) {/eq}. In this problem, we use {eq}\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1} {/eq}.

Answer and Explanation:

1. {eq}\displaystyle d\left(\frac{\cos x }{x^2}\right)\\ f(x) = \cos x , g(x) = x^2\\ \displaystyle d\left(\frac{\cos x }{x^2}\right) \displaystyle = \frac{x^2 d( \cos x) - \cos x d(x^2)}{(x^2)^2}\ \ \ \ \ (By \ Quotient \ Rule)\\ \displaystyle d\left(\frac{\cos x }{x^2}\right) \displaystyle = \frac{- x^2 \sin x -2x \cos x}{x^4}\\ \displaystyle d\left(\frac{\cos x }{x^2}\right) = \frac{-x \sin x -2 \cos x}{x^3}\\ {/eq}

2.{eq}\displaystyle \int \dfrac {6 x^5 \ dx} {(8 + x^6)^5}\\ put \ 8 + x^6 = y {/eq}

Differentiating this with respect to {eq}x {/eq} we get {eq}6x^5 dx = dy \\ {/eq}

Therefore {eq}\displaystyle \int \dfrac {dy} {(y)^5} = \displaystyle \int y^{-5} dy = \frac{y^{-4}}{-4} +c \\ {/eq}

Therefore {eq}\displaystyle \int \dfrac {6 x^5 \ dx} {(8 + x^6)^5} = \displaystyle \frac{y^{-4}}{-4} +c \\ \displaystyle \int \dfrac {6 x^5 \ dx} {(8 + x^6)^5} = \frac{(8 + x^6)^{-4}}{-4} +c {/eq}.


Learn more about this topic:

Loading...
Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
112K

Related to this Question

Explore our homework questions and answers library