Find the derivative of the following?

Question

Find:
d/dx [(x² + √x - 4)/(√x³ - ³√x + π)]

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...easy...
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Answer 1

³ ² √ π

Answer!!

= (2x√ x³) - (2³√x) + (2xπ)+ (√x³ / 2√x )- (³√x / 2√ x) + ( π / 2√x) - (3x²√x / 2) -(3x / 2) + (6√x) - (x² / 3³√x²) - (√x / 3³√x²) + (4 / 3³√x²)


ALL OVER!!!

(√x³ - ³√x + π)²


h3h3h3 that's all pretty long!!

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Answer 2

Solution:
Let u(x) = x^2 + sqrt(x) - 4 = x^2 + x^(1/2) - 4
and v(x) = sqrt(x^3) - cbrt(x) + pi = x^(3/2) - x^(1/3) + pi.

Then,
u'(x) = 2x + (1/2) x^(-1/2)
= (1/2)x^(-1/2) [4x^(3/2) + 1].

Similarly,
v'(x) = (3/2)x^(1/2) - (1/3)x^(-2/3)
= (3/2)x^(3/6) - (1/3)x^(-4/6)
= (1/3)x^(-4/6) [(9/2)x^(7/6) - 1]
= (1/3)x^(-2/3) [(9/2)x^(7/6) - 1].

Thus,

d/dx [u(x) / v(x) ] = [ v(x) u'(x) - u(x) v'(x) ] / [v(x)]^2
= (1/2)x^(-1/3) [x^(3/2) - x^(1/3) + pi][4x^(9/6) + 1] / [v(x)]^2
- (1/3)x^(-2/3) [x^2 + x^(1/2) - 4][(9/2)x^(7/6) - 1] / [v(x)]^2,

where [v(x)]^2 = [x^(3/2) - x^(1/3) + pi]^2.

Answer 3

I don't see any tricks to this. It is possible (except where the denominator is 0, if that occurs, and for x<4) but it'll be messy. You need to apply either the quotient rule, or the product rule.

Answer 4

lol