# differentiate with respect to x

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• Nov 16th 2007, 02:02 PM
orsonik
differentiate with respect to x

(a) y = e^x ln x + x^3
Hint: Note that the first term on the right hand side is a
product of functions.
(b) y = e^2x^2
Hint: Use the chain–rule with u = 2x^2.

I have no clue how to solve these.help please is there a special equotation I have to use????

• Nov 16th 2007, 02:42 PM
topsquark
Quote:

Originally Posted by orsonik

(a) y = e^x ln x + x^3
Hint: Note that the first term on the right hand side is a
product of functions.
(b) y = e^2x^2
Hint: Use the chain–rule with u = 2x^2.

I have no clue how to solve these.help please is there a special equotation I have to use????

First note that
$\displaystyle \frac{d}{dx}e^x = e^x$

$\displaystyle \frac{d}{dx}ln(x) = \frac{1}{x}$

So
$\displaystyle y = e^x~ln(x) + x^3$

$\displaystyle \frac{dy}{dx} = \left ( \frac{d}{dx} e^x \right ) ln(x) + e^x \frac{d}{dx}ln(x) + \frac{d}{dx}x^3$

$\displaystyle \frac{dy}{dx} = e^x~ln(x) + e^x \cdot \frac{1}{x} + 3x^2$

For the second one:
$\displaystyle y = e^{2x^2}$

Let $\displaystyle f(u) = e^u$ and $\displaystyle u(x) = 2x^2$.

So
$\displaystyle y = f(u(x))$

$\displaystyle \frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$

$\displaystyle \frac{dy}{dx} = e^{u(x)} \cdot 4x$

$\displaystyle \frac{dy}{dx} = 4x~e^{2x^2}$

-Dan
• Nov 16th 2007, 02:46 PM
orsonik
yyyy
oh my days. you are a real matematician magitian. I wish I was as good but I will probably never understand maths:(((:confused: