How do you find the derivative of #y=x^nlnx#?

Question

Isaiah Allen · Accepted Answer

# d/dx(x^nlnx)=x^(n-1) + nx^(n-1)lnx #

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

# d/dx(uv)=u(dv)/dx+(du)/dxv #, or, # (uv)' = (du)v + u(dv) #

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

This can be extended to three products:

# d/dx(uvw)=uv(dw)/dx+u(dv)/dxw + (du)/dxvw#

So with # f(x) = x^2sinxtanx # we have;

# { ("Let "u = x^n, => , (du)/dx = nx^(n-1)'), ("And "v = lnx, =>, (dv)/dx = 1/x' ) :}#

Applying the product rule we get:

# \ \ \ \ \ \ \ \ \ \ \ d/dx(uv)=u(dv)/dx + (du)/dxv # # :. d/dx(x^nlnx)=(x^n)(1/x) + (nx^(n-1))(lnx) # # :. d/dx(x^nlnx)=x^(n-1) + nx^(n-1)lnx #

Cameron Alexander · Answer

undefined To find the derivative of $ y = x^n \ln x $, you can use the product rule. The product rule states that if $ u $ and $ v $ are both functions of $ x $, then the derivative of their product is given by $ (uv)' = u'v + uv' $. Applying this rule to the function $ y = x^n \ln x $, we get:

$$ y' = (x^n)' \ln x + x^n (\ln x)' $$

Now, differentiate each term separately using the power rule and the derivative of natural logarithm, which is $ (\ln x)' = \frac{1}{x} $:

$$ y' = (n x^{n-1}) \ln x + x^n \cdot \frac{1}{x} $$

$$ y' = n x^{n-1} \ln x + \frac{x^n}{x} $$

$$ y' = n x^{n-1} \ln x + x^{n-1} $$

Therefore, the derivative of $ y = x^n \ln x $ with respect to $ x $ is $ y' = n x^{n-1} \ln x + x^{n-1} $.