# PracticeDifferentiate 3

Banned
Differentiate.

#### TheJason

Staff member
Moderator
$$\displaystyle y = \dfrac{2x - \cos\,x }{x^{2}}$$

$$\displaystyle y'= \dfrac{(x^{2})(2 - (-\sin\,x)) - (2x)(2x - \cos\,x) }{(x^{2})^{2}}$$

$$\displaystyle y'= \dfrac{(x^{2})(2 + \sin\,x) - (2x)(2x - \cos\,x) }{(x^{2})^{2}}$$

$$\displaystyle y' = \dfrac{2x^{2} + \sin^{3}x - 4x^{2} - 2\cos^{2}x}{x^{4}}$$

$$\displaystyle y' = \dfrac{-2x^{2} + \sin^{3}x - 2\cos^{2}x}{x^{4}}$$

Last edited:

#### MarkFL

##### La Villa Strangiato
Staff member
Moderator
Math Helper

$$\displaystyle x^2\cdot\sin(x)\ne\sin^3(x)$$

$$\displaystyle x\cdot\cos(x)\ne\cos^2(x)$$

anemone

#### TheJason

Staff member
Moderator

$$\displaystyle x^2\cdot\sin(x)\ne\sin^3(x)$$

$$\displaystyle x\cdot\cos(x)\ne\cos^2(x)$$
$$\displaystyle y = \dfrac{2x - \cos\,x }{x^{2}}$$

$$\displaystyle y'= \dfrac{(x^{2})(2 - (-\sin\,x)) - (2x)(2x - \cos\,x) }{(x^{2})^{2}}$$

$$\displaystyle y'= \dfrac{(x^{2})(2 + \sin\,x) - (2x)(2x - \cos\,x) }{(x^{2})^{2}}$$

$$\displaystyle y'= \dfrac{2x^{2} + x^{2}\sin\,x - 4x^{2} + 2x\cos\,x }{x^{4}}$$

$$\displaystyle y'= \dfrac{-2x^{2} + x^{2}\sin\,x + 2x\cos\,x }{x^{4}}$$

#### MarkFL

##### La Villa Strangiato
Staff member
Yes, that appear to be correct...I would go on to divide by numerator and denominator by $$x$$.