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)

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.

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

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

In plain text, is written x^y. Also, it is important to write parentheses because is completely different from . 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.