Given $y = e^((ln x)^2)$ how do you find find y'(e)?

Question

26156 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-02T20:50:09

$dy/dx = (2e^(ln^2x)lnx)/x => y'(e) = 2$

We have:

$y = e^(ln^2x)$

Take Natural Logarithms of both sides:

$ln y = (lnx)^2$

Differentiate Implicitly, and apply the chain rule:

$1/y * dy/dx = 2(lnx)*(1/x)$

Which we can rearrange to get:

$dy/dx = (2ylnx)/x$
$" " = (2e^(ln^2x)lnx)/x$

So then, when $x=e$ we have:

$dy/dx = (2e^(ln^2e)lne)/e$
$" " = (2e^1*1)/e$
$" " = 2$

Answer 2 · 2017-05-02T22:33:07

Recall that $d/dxe^x=e^x$ . Then, through the chain rule, $d/dxe^f(x)=e^f(x)f'(x)$ .

So:

$y(x)=e^((lnx)^2)$

$y'(x)=e^((lnx)^2)d/dx(lnx)^2$

$color(white)(y'(x))=e^((lnx)^2)(2(lnx))d/dxlnx$

$color(white)(y'(x))=(2e^((lnx)^2)lnx)/x$

Then:

$y'(e)=(2e^((lne)^2)lne)/e$

$color(white)(y'(e))=(2e^1(1))/e$

$color(white)(y'(e))=2$

Given y = e^((ln x)^2)y = e^((ln x)^2) how do you find find y'(e)?