How to differentiate y=e^(xlnx)?

Question

anonymous · Answer

You have to use the chain rule.

anonymous · Answer

$$
y = e^u\implies \frac{d}{dx} y = \frac d{dx}\bigg(u\bigg)e^u = \frac d{dx}\bigg(x\ln x\bigg)e^{x\ln x}
$$

solomonzelman · Answer

I am sure using logarithmic differentiation is a bit better approach.
but you can follow wio, as he knows much more than me.

anonymous · Answer

u=lnx
du/dx=lnx+1
dy/dx=dy/du.du/dx
            e^u .(lnx+1)
            e^(xlnx) .(lnx+1)
is this true?

solomonzelman · Answer

yes.

solomonzelman · Answer

you differentiated it correctly
So,     $\large\color{black}{ y'=e^{x \ln(x) } ~^{.} (\ln(x)+1)   }$

anonymous · Answer

One thing to note:$$
e^{x\ln x} = \big(e^{\ln x}\big)^x  =x^x
$$

anonymous · Answer

i want to know how to slove this by logarithmic differentiation .can you explane it?

anonymous · Answer

What is being ton here is logarithmic differentiation, in a way.

anonymous · Answer

thaks you!

solomonzelman · Answer

$\large\color{black}{ y=e^{x \ln(x) }   }$ 

hit both sides with ln.

$\large\color{black}{ \ln(y)=x \ln(x) ~\ln(e)   }$ 

$\large\color{black}{ \ln(y)=x \ln(x)    }$

solomonzelman · Answer

then differentiate both sides

anonymous · Answer

For logarithmic differentiation, you use: $$
\ln(y) = x\ln x
$$Differentiate both sides: $$
\frac{y'}{y}=\ln x+1
$$

anonymous · Answer

Then substitute: $$
\frac{y'}{e^{x\ln x}}=\ln x+1
$$

solomonzelman · Answer

yes, that is the derivative. the derivative of ln(y) is 1/y, and we multiply times y'.

solomonzelman · Answer

and the right side (as you have already differentiated) went by the product rule.

solomonzelman · Answer

$\large\color{black}{      \frac{\LARGE y'}{\LARGE y}=\ln(x)+1               }$

solomonzelman · Answer

you multiply both sides times, y, and since we know from the beginning that:
$\large\color{black}{      y=e^{x \ln(x)}               }$ , therefore, you substitute, $\large\color{black}{      e^{x \ln(x)}               }$  instead of $\large\color{black}{      y           }$ .

solomonzelman · Answer

you will then get (still) the same answer.

solomonzelman · Answer

actually chain is better, sorry..

anonymous · Answer

itz ok thanks alot

solomonzelman · Answer

yw

anonymous · Answer

in sort: $$
y=f(x)\
\ln(y)=\ln(f(x))\
y'=y(\ln(f(x)))'
$$We do this whenever $\ln(f(x))$ is easier to differentiate than $f(x)$.

anonymous · Answer

This is the same as doing: $$
y=f(x)\
y=e^{\ln (f(x))} \
y' = (\ln(f(x)))'e^{\ln(f(x))}\
y'= (\ln(f(x)))'f(x)
$$

solomonzelman · Answer

a couple of examples would look easier.... if there is anything you don't get ask.

anonymous · Answer

ok sure thanks alot again

solomonzelman · Answer

you welcome~!