How do you differentiate #y=(x^2lnx)^4#?

Question

Scarlett Allison · Accepted Answer

# dy/dx = 4x(x^2lnx)^3 ( 1 + 2lnx ) # # y = (x^2lnx)^4 # By the chain rule, # d/dxf(g(x)) =f'(g(x))g'(x)# or,# dy/dx=dy/(du)(du)/dx #, we have: # dy/dx = 4(x^2lnx)^3 d/dx(x^2lnx)# Next, we need to use the product rule; # d/dx(uv)=u(dv)/dx+v(du)/dx # to get: # dy/dx = 4(x^2lnx)^3 { (x^2)(d/dxlnx) + (d/dxx^2)(lnx) } # # :. dy/dx = 4(x^2lnx)^3 { (x^2)(1/x) + (2x)(lnx) } # # :. dy/dx = 4(x^2lnx)^3 ( x + 2xlnx ) # # :. dy/dx = 4x(x^2lnx)^3 ( 1 + 2lnx ) #

Axel Alvarez · Answer

undefined To differentiate $ y = (x^2 \ln x)^4 $, you would use the chain rule. Here's the step-by-step process:

1. Let $ u = x^2 \ln x $.
2. Differentiate $ u $ with respect to $ x $ to find $ \frac{du}{dx} $.
3. Apply the power rule to $ u^4 $ to find $ \frac{dy}{du} $.
4. Multiply $ \frac{du}{dx} $ and $ \frac{dy}{du} $ to find $ \frac{dy}{dx} $, using the chain rule.

Here's the breakdown:

1. $ u = x^2 \ln x $
   - $ \frac{du}{dx} = 2x \ln x + x $
2. $ \frac{dy}{du} = 4u^3 $
   - $ \frac{dy}{du} = 4(x^2 \ln x)^3 $
3. $ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} $
   - $ \frac{dy}{dx} = 4(x^2 \ln x)^3 \times (2x \ln x + x) $

So, the derivative of $ y = (x^2 \ln x)^4 $ with respect to $ x $ is:

$$ \frac{dy}{dx} = 4(x^2 \ln x)^3 \times (2x \ln x + x) $$