Find y'.
-$$y = \left( \frac { 1 } { x } + 7 \right) \left( x - \frac { 1 } { x } + 7 \right)$$
A) $- \frac { 1 } { x ^ { 3 } } - 7$
B) $\frac { 1 } { x ^ { 3 } } + 7$
C) $\frac { 2 } { x ^ { 3 } } + 7$
D) $- \frac { 2 } { x ^ { 3 } } - 7$

Question

Quizplus · Accepted Answer

Using the product rule, we have:

$$y'=\left(\frac{-1}{x^2}+0\right)\left(x-\frac{1}{x}+7\right)+\left(\frac{1}{x}+7\right)\left(1+\frac{1}{x^2}\right)$$
Simplifying, we get:

$$y'=-\frac{1}{x^2}\left(x-\frac{1}{x}+7\right)+\frac{1}{x}+7+\frac{1}{x^2}\left(x-\frac{1}{x}+7\right)$$
$$y'=-\frac{1}{x^2}+\frac{1}{x}+7+\frac{1}{x^2}+\frac{7}{x}+\frac{7}{x^2}$$
$$y'=\frac{2}{x^3}+7$$

Therefore, the answer is (C).

Find Y' - y=(1x+7)(x−1x+7)y = \left( \frac { 1 } { x } + 7 \right) \left( x - \frac { 1 } { x } + 7 \right)y=(x1​+7)(x−x1​+7)

Find Y'
- $y = \left( \frac { 1 } { x } + 7 \right) \left( x - \frac { 1 } { x } + 7 \right)$