1. ## Implicit Differentiation

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

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

2. I assume you are taking the derivative with respect to x? If so you need to use the chain rule, i.e. $\frac{d}{dx} (f\circ g)(x) = (f'\circ g)(x)\cdot g'(x)$.

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

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

I hope that helps.

3. Originally Posted by lvleph

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

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

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

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

this :

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

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

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

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

sorry i don't have intention to be rude or something but to do derivation of function and constant like derivation of two functions ... that's not wrong but not needed at all

4. Originally Posted by yeKciM
$\displaystyle \frac {\delta}{\delta x } (xy^2) = y^2$

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

$\displaystyle \frac {\delta}{\delta x } (xy^2) = y^2 \cdot \frac {\delta}{\delta x } (x) = y^2$
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 $y'=0$ and everything simplifies to your case.

5. Originally Posted by lvleph
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 $y'=0$ and everything simplifies to your case.
yes but everything that is not "x" when you do derivation on x is constant

6. No, it is not. Here is a good example. $\frac{d}{dx}\left[x\cdot f^2(x)\right] = x\cdot2f(x) \cdot f'(x) + f(x)$.

7. Originally Posted by lvleph
No, it is not. Here is a good example. $\frac{d}{dx}\left[x\cdot f^2(x)\right] = x\cdot2f(x) \cdot f'(x) + f(x)$.
yes but, that (f(x)) is function (or any mapping) of x. or am I mistaking ? it's not like function of y or anything else

sorry, but now i'm like confused

Edit: aaaah sorry I totally forgot about question up there... I'm very very sorry

8. Originally Posted by Sangreal
How can I 'evaluate' the following derivative and simplify it ?

xy^2 = 3^(y/ln x) + sin^2 x
Differentiating each term independently...

LHS

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

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

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

RHS

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

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

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

giving

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

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

Differentiating $Sin^2(x)$

let $u=Sin(x)$

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

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

so you will need to factor it out...