How do you use the chain rule to differentiate #y=root4(-3x^4-2)#?

Question

Emma Aldridge · Accepted Answer

#d/dxroot(4)(-3x^4-2)=(-3x^3)/root(4)((-3x^4-2)^3)#

First recognize that the chain-rule means if you have a function within a function, you derive the "outer" function and then the "inner" function. To put this in a mathematical form:

#d/dxf(g(x))=f'(g(x))*g'(x)#

So first, let's look at #root(4)(-3x^4-2)#. Notice that we can write this in its exponent form to get a clear image: #(-3x^4-2)^(1/4)#.

First we derive the outside function or the exponent in this case. So we get #1/(4(-3x^4-2)^(3/4))#.

Then we have to recognize that there's a function within the root: #(-3x^4-2)#. Deriving this, we get #(-12x^3)#.

Then using the chain-rule as a guide, we will multiply #g'(x)# or #(-12x^3)# in this case with #f'(x)#:

#1/(4(-3x^4-2)^(3/4))*(-12x^3)#

Notice we can simplify the constant #-12/4# to #-3#.

So our final answer is: #(-3x^3)/root(4)((-3x^4-2)^3)#

Note: you can write: #root(4)((-3x^4-2)^3)# as #(-3x^4-2)^(3/4)# for the final answer in the denominator.

Emily Ammerman · Answer

undefined To differentiate $y = \sqrt[4]{-3x^4 - 2}$ using the chain rule, first rewrite the function in a more suitable form:

$y = (-3x^4 - 2)^{1/4}$

Then, differentiate using the chain rule:

$y' = \frac{1}{4}(-3x^4 - 2)^{-3/4} \cdot (-12x^3)$

Simplify the expression:

$y' = -3x^3(-3x^4 - 2)^{-3/4}$

This is the derivative of the given function.