# Thread: Am I on the right track with quotient rule?

1. ## Am I on the right track with quotient rule?

Differentiate with respect to x using the quotient rule?
Y= (2x4 – 3x) / (4x -1)

d/dx [2x4 – 3x] = 8x3 – 3
and
d/dx [4x – ] = 4

by the quotient rule

f(x) = (g(x)) / (h(x)) , f’(x) = [g1(x) h(x) – g(x)h1(x)] / [h(x)]2

choosing g(x) = 2x4 – 3x and h(x) = 4x - 1

gives

dy/dx = [(8x3 – 3) (4x – 1) - (2x4 – 3x) (4) ] / (4x – 1) 2
therefore

dy/dx = (23x4 – 8x3 – 12x + 3 – 8x4 + 12x) / 16x2 – 8x + 1

therefore
dy/dx = (24x4 – 8x3 + 3) / (16x2 – 8x + 1)

2. ## Re: Am I on the right track with quotient rule?

In the second last line, it is not clear where 23x^4 appeared from. Also, in the same line, division is done before subtraction, which is not what you want. The final result is correct.

3. ## Re: Am I on the right track with quotient rule?

sorry a typo there was meant to be 32.

could I check another one please

Find the integral:

∫ (5x2 + √x- (4/x2)) dx

= (5x3/3) + (2x3/2/3) + 4/x + c

= ((5x3 + 2x3/2 ) / 3) + (4/x) + c , where c is a constant

Or

= ((5x4 + 2x5/2 + 12) / 3x )) + c , where c is a constant

4. ## Re: Am I on the right track with quotient rule?

Find the integral:
∫ (5x2 + √x- (4/x2)) dx
= (5x3/3) + (2x 3/2/3) + 4/x + c
= ((5x3 + 2x 3/2 ) / 3) + (4/x) + c , where c is a constant
Or
= ((5x4 + 2x 5/2 + 12) / 3x )) + c , where c is a constant

5. ## Re: Am I on the right track with quotient rule?

In plain text, $x^y$ is written x^y. Also, it is important to write parentheses because $2x^{3/2}$ is completely different from $\frac{2x^3}{2}$. You could check the result yourself in WoldramAlpha. Just don't overuse it and do the work by hand first if you are supposed to.