How do you find the derivative of #g(x)=(2x^2+x+1)^-3#?

Question

Cameron Alexander · Accepted Answer

#-3(4x+1)(2x^2+2x+1)^-3#

We use the chain rule, which states that,

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

Let #u=2x^2+x+1,:.(du)/dx=4x+1#.

Now, #y=u^-3,:.dy/(du)=-3u^-4#.

And so,

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

Now, substitute back #u=2x^2+x+1# to get:

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

I'd clean this up and rearrange it into:

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

Emma Aldridge · Answer

undefined To find the derivative of $ g(x) = (2x^2 + x + 1)^{-3} $, you can use the chain rule.

1. Start by applying the chain rule, which states that if $ u $ is a function of $ x $ and $ f(u) $ is a function of $ u $, then the derivative of $ f(u) $ with respect to $ x $ is $ \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} $.
2. Let $ u = 2x^2 + x + 1 $.
3. Find the derivative of $ u $ with respect to $ x $, which is $ \frac{du}{dx} = 4x + 1 $.
4. Find the derivative of $ g(u) = u^{-3} $ with respect to $ u $, which is $ \frac{dg}{du} = -3u^{-4} $.
5. Multiply the derivatives from steps 3 and 4 together to get the derivative of the original function.
6. Substitute $ u = 2x^2 + x + 1 $ back into the result.

Therefore, the derivative of $ g(x) $ with respect to $ x $ is $ g'(x) = -3(2x^2 + x + 1)^{-4} \cdot (4x + 1) $.