How do you differentiate #(3x-4/x)^2 (1-x+7x^2)^4#?

Question

Emilia Aguilar · Accepted Answer

Chain rule and product rule time combined!

Let's work with your function #f(x) = (3x-4x^-1)^2(1-x+7x^2)^4#.

First, we will rename:

#u=(3x-4x^-1)^2# and #w=(1-x+7x^2)^4#

Using the product rule, as follows,

#(d(uw))/(dx)=w((du)/(dx))+u((dw)/(dx))#

#(1-x+7x^2)^4*color(green)((d(3x-4x^-1)^2)/dx) + (3x-4x^-1)^2*color(blue)((d(1-x+7x^2)^4)/dx)#

Now, we will lead with the colored derivatives first and then return to the whole equation above.

To solve the green derivative, let's name #a=3x-4x^-1# and solve it as a chain rule, where

#(dy)/(da)*(da)/(dx)=(dy)/(dx)#

#(dy)/(dx)=2(3x-4x^-1)(3+4x^-2)#

The same for the blue part, where #b=1-x+7x^2#:

#(dy)/(dx)=4(1-x+7x^2)^3(-1+14x)#

Now we have the derivatives #(du)/(dx)# and #(dw)/(dx)#, let's go back to the partillay colored long equation:

#(1-x+7x^2)^4*2(3x-4x^-1)*(3+4x^-2) + (3x-4x^-1)^2*4(1-x+7x^2)^3*(-1+14x)#

You may want to expand it and eliminate the factoring (parenthesis), but that is not really necessary as the work and steps to achieve so would not necessarily mean simplification. The line above can be your answer.

Isabella Ackerman · Answer

undefined To differentiate the expression $ (3x - \frac{4}{x})^2 (1 - x + 7x^2)^4 $, you can use the product rule. The product rule states that if you have two functions, $ u $ and $ v $, then the derivative of their product is given by $ u'v + uv' $. Here's how you apply it to this expression:

Let $ u = (3x - \frac{4}{x})^2 $ and $ v = (1 - x + 7x^2)^4 $.

Find $ u' $ and $ v' $, then apply the product rule:

$$ u' = 2(3x - \frac{4}{x}) \left(3 - \frac{-4}{x^2}\right) $$
$$ v' = 4(1 - x + 7x^2)^3(-1 + 14x) $$

Now, apply the product rule:

$$ \frac{d}{dx}[(3x - \frac{4}{x})^2 (1 - x + 7x^2)^4] = u'v + uv' $$

$$ = (2(3x - \frac{4}{x}) \left(3 - \frac{-4}{x^2}\right))(1 - x + 7x^2)^4 + (3x - \frac{4}{x})^2(4(1 - x + 7x^2)^3(-1 + 14x)) $$

So, the derivative of $ (3x - \frac{4}{x})^2 (1 - x + 7x^2)^4 $ is the expression above.