How can I 'evaluate' the following derivative and simplify it ?

xy^2 = 3^(y/ln x) + sin^2 x

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- Aug 24th 2010, 05:24 AMSangrealImplicit Differentiation
How can I 'evaluate' the following derivative and simplify it ?

xy^2 = 3^(y/ln x) + sin^2 x - Aug 24th 2010, 09:09 AMlvleph
I assume you are taking the derivative with respect to x? If so you need to use the chain rule, i.e. $\displaystyle \frac{d}{dx} (f\circ g)(x) = (f'\circ g)(x)\cdot g'(x)$.

To start you off

$\displaystyle \frac{d}{dx}\left[xy^2\right] = y^2 + x\cdot 2y\cdot y'$.

Also, if $\displaystyle f(x) = a^{g(x)}$ then

$\displaystyle \frac{d}{dx} f(x) = \frac{d}{dx} a^{g(x)} \Rightarrow \frac{d}{dx} \left[\ln f(x)\right] = \frac{d}{dx} \left[g(x)\cdot \ln a\right] \Rightarrow \frac{1}{f(x)}\cdot f'(x) = \ln a \cdot g'(x) \Rightarrow f'(x) = f(x) \cdot g'(x) \ln a$

I hope that helps. - Aug 24th 2010, 09:21 AMyeKciM

$\displaystyle \displaystyle \frac {d}{d x } (xy^2) = y^2 $

you don't need to do it like that... (i assume that your last " y' " is derivated by the x :D ) it's much easier (for more complex problems) to just get the constant (in this case y) out ...

$\displaystyle \displaystyle \frac {d}{d x } (xy^2) = y^2 \cdot [ \frac {d}{d x } (x) ]= y^2$

this :

$\displaystyle (u\cdot v)' = u'v +v'u $

is being used for derivate something like this (where are two functions )

$\displaystyle \displaystyle \frac {d}{d x } (e^x \cdot x^2 ) $

that what you did there is like that for (2x^2) for derivate on "x" you do $\displaystyle (2x^2)'= 2' \cdot x^2 + 2 \cdot (x^2)' = 0\cdot x^2 + 2\cdot 2x = 4x$ :D:D:D:D:D that's to much work to be done, especially when we have constant on our hands :D:D:D:D

sorry i don't have intention to be rude or something :D but to do derivation of function and constant like derivation of two functions ... that's not wrong but not needed at all :D - Aug 24th 2010, 09:26 AMlvleph
Except that you are assuming that y is a constant with respect to x and I was not assuming this. Because there was no statement of dependence of y on x it is more correct to assume dependence and then if there is no dependence we can just say $\displaystyle y'=0$ and everything simplifies to your case.

- Aug 24th 2010, 09:43 AMyeKciM
- Aug 24th 2010, 09:46 AMlvleph
No, it is not. Here is a good example. $\displaystyle \frac{d}{dx}\left[x\cdot f^2(x)\right] = x\cdot2f(x) \cdot f'(x) + f(x)$.

- Aug 24th 2010, 10:02 AMyeKciM
- Aug 24th 2010, 10:11 AMArchie Meade
Differentiating each term independently...

**LHS**

$\displaystyle xy^2\;\;\;$ is a product.

$\displaystyle \displaystyle\frac{d}{dx}\left(xy^2\right)=y^2\fra c{d}{dx}x+x\frac{d}{dx}{y^2}=y^2+x\frac{dy}{dx}\ \frac{d}{dy}y^2$

$\displaystyle =\displaystyle\ y^2+2xy\frac{dy}{dx}$

**RHS**

$\displaystyle \displaystyle\huge\ 3^{\left(\frac{y}{lnx}\right)}=3^{h(x)}$

$\displaystyle \displaystyle\frac{d}{dx}3^{h(x)}=\frac{d}{dh(x)}3 ^{h(x)}\frac{d}{dx}h(x)=3^{h(x)}ln3\frac{d}{dx}h(x )$

since $\displaystyle \displaystyle\frac{d}{dx}a^x=a^xlna$

giving

$\displaystyle \displaystyle\frac{d}{dx}3^{\left(\frac{y}{lnx}\ri ght)}=3^{\left(\frac{y}{lnx}\right)}ln3\frac{d}{dx }\left(\frac{y}{lnx}\right)$

$\displaystyle \displaystyle\frac{d}{dx}\left(\frac{y}{lnx}\right )=\frac{lnx\frac{dy}{dx}-y\left(\frac{1}{x}\right)}{(lnx)^2}$

Differentiating $\displaystyle Sin^2(x)$

let $\displaystyle u=Sin(x)$

$\displaystyle \displaystyle\frac{d}{du}u^2\frac{du}{dx}=2uCosx=2 SinxCosx=Sin(2x)$

You need to get $\displaystyle \displaystyle\frac{dy}{dx}=........$ as $\displaystyle \displaystyle\frac{dy}{dx}$ occurs more than one when you write LHS=RHS.

so you will need to factor it out...