How do you differentiate # 3/4 * (2x^3 + 3x)^(-1/4)#?

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Answer 1 · 2016-06-22T15:27:46

#(-18x^2+9)/(16(2x^3+3x)^(5/4)#

We can factor a constant out of the equation:
#3/4 * [(2x^3+3x)^(-1/4)]#

Then we apply the chain rule: #d/dx[f(g(x))] -> f'(g(x)) * g'(x)#

#d/dx[(2x^3+3x)^(-1/4)] -> -1/4(2x^3+3x)^(-5/4) #

#d/dx[2x^3+3x] -> 6x^2+3#

We multiply the outside derivative and inside derivative together:
#3/4 * [-1/4(2x^3+3x)^(-5/4) * (6x^2+3)]#

Bring the negative exponent down below:
#3/4*-(6x^2+3)/(4(2x^3+3x)^(5/4)#

Multiply the constant across:
#(-18x^2+9)/(16(2x^3+3x)^(5/4)#