# Derivative

• Sep 29th 2009, 08:15 PM
Derivative
f(x)= 10x^(lnx)
How do you find the value of the derivative when x =8?
• Sep 29th 2009, 08:31 PM
Roam
Quote:

Originally Posted by LinaD3
f(x)= 10x^(lnx)
How do you find the value of the derivative when x =8?

First you need to differentiate it. Have you done this? Then substitute x = 8 into f'(x).
• Sep 29th 2009, 08:32 PM
But I'm supposed to find the value of 8 for the derivative of that function. How do I take the derivative of it?
• Sep 29th 2009, 08:42 PM
Roam
You have: $f(x) = 10x^{ln(x)}$

Apply the chain rule:

$
10x^{ln (x)} \frac{d}{dx} (ln^2x)
$

u=log x and n=2

$
= 20x^{ln(x)} ln(x) \frac{d}{dx} (ln(x))
$
and since the derivative of ln(x) = 1/x

you have: $f'(x) = 20 x^{ln(x)-1}ln(x)$
• Sep 29th 2009, 08:43 PM
I don't know how to differentiate it. Can you give me a hint on how to?
• Sep 29th 2009, 08:48 PM
Roam
Quote:

Originally Posted by LinaD3
I don't know how to differentiate it. Can you give me a hint on how to?

You have to use the chain rule. I already did it for you in my last post. Now all you need to do is to substitute $x=8$
• Sep 30th 2009, 01:52 AM
mr fantastic
Quote:

Originally Posted by Roam
You have: $f(x) = 10x^{ln(x)}$

Apply the chain rule:

$
10x^{ln (x)} \frac{d}{dx} (ln^2x)
$

u=log x and n=2

$
= 20x^{ln(x)} ln(x) \frac{d}{dx} (ln(x))
$
and since the derivative of ln(x) = 1/x

you have: $f'(x) = 20 x^{ln(x)-1}ln(x)$

Your answer is correct but I cannot follow the logic.

Using logarithmic differentiation:

$y = 10 x^{\ln x} \Rightarrow \ln y = (\ln x)^2 + \ln 10$.

Therefore $\frac{1}{y} \frac{dy}{dx} = \frac{2 \ln x}{x} \Rightarrow \frac{dy}{dx} = \frac{2 y \ln x}{x} = \frac{2 \left(10 x^{\ln x}\right) \ln x}{x} = 20 x^{\ln (x) - 1} \ln x$.
• Sep 30th 2009, 08:50 AM
tom@ballooncalculus
Also, just in case a picture helps convey the logic that Roam (possibly) and - Wolfram|Alpha (definitely) are using...

http://www.ballooncalculus.org/asy/diffMulti/one.png

... where

http://www.ballooncalculus.org/asy/doubleChain.png

... is the chain rule for two inner functions, i.e...

$\frac{d}{dx}\ f(u(x), v(x)) = \frac{\partial f}{\partial u} \frac{du}{dx} + \frac{\partial f}{\partial v} \frac{dv}{dx}$

As with...

http://www.ballooncalculus.org/asy/chain.png

... the ordinary chain rule, straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the (corresponding) dashed balloon expression which is (one of) the inner function(s) of the composite expression.

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