Differentiate y=(4x+x^-5)^1/3

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given y= (4x+x^-5)^1/3

We need to differentiate.

We will use the chain rule to differentiate.

==> Let y= u^1/3

==> u= 4x+x^-5  ==>  u' = 4 - 5x^-6

Then we know that:

y' = 1/3 * u^-2/3  * u'

Let us substitute.

==> y' = (1/3) * (4x+x^-5)^-2/3 * (4-5x^-6)

==> y' =  (1/3)*(4-5x^-6)/ (4x+x^-5)^2/3

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

This problem requires the use of chain rule since y is the result of composition of 2 functions u = 4x+x^-5 and v = u^(1/3).

We'll differentiate first the power, what's inside brackets remaining unchanged, then we'll differentiate what's inside brackets, with respect to x.

dy/dx = (1/3)*(4x+x^-5)^(1/3 - 1)*(4x+x^-5)'

dy/dx = (1/3)*(4x+x^-5)^(-2/3)*(4 + -5*x^(-6))

dy/dx = (4 - 5/x^6)/3*(4x + 1/x^5)^(2/3)

dy/dx = [(4x^6 - 5)/x^6]/3*[(4x^6+1)/x^5]^(2/3)

The result of differentiating y is:

dy/dx = [(4x^6 - 5)/x^6]/3*[(4x^6+1)/x^5]^(2/3)

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