What is the derivative of #y=(x^2lnx)^4#?

Question

Isaiah Allen · Accepted Answer

#y^' = 4x^7 * ln^3(x) * (2ln(x) + 1)#

Notice that you can simplify your function to get

#y = (x^2)^4 * (lnx)^4#

#y = x^8 * ln^4(x)#

To differentiate this function you can use the product rule and the chain rule for #u^4#, with #u = ln(x)#.

#d/dx(y) = [d/dx(x^8)] * ln^4(x) + x^8 * d/dx(ln^4(x))#

#y^' = 8x^7 * ln^4(x) + x^8 * ([d/(du)(u^4)] * d/dx(u))#

#y^' = 8x^7 * ln^4(x) + x^color(red)(cancel(color(black)(8))) * 4 ln^3(x) * 1/color(red)(cancel(color(black)(x)))#

#y^' = 8x^7 * ln^4(x) + 4x^7 * ln^3(x)#

This can be simplified to give

#y^' = color(green)(4x^7 * ln^3(x) * (2ln(x) + 1))#

Elijah Abram · Answer

undefined To find the derivative of $ y = (x^2 \ln x)^4 $, you can use the chain rule. The derivative is:

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

Using the product rule to differentiate $x^2 \ln x$, you get:

$$ \frac{d}{dx}(x^2 \ln x) = \frac{d}{dx}(x^2) \cdot \ln x + x^2 \cdot \frac{d}{dx}(\ln x) $$

Simplify and differentiate:

$$ = 2x \ln x + x^2 \cdot \frac{1}{x} $$

$$ = 2x \ln x + x $$

Now substitute this back into the original expression:

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

This is the derivative of $ y = (x^2 \ln x)^4 $.