# Implicit Differentiation

• Aug 24th 2010, 06:24 AM
Sangreal
Implicit Differentiation
How can I 'evaluate' the following derivative and simplify it ?

xy^2 = 3^(y/ln x) + sin^2 x
• Aug 24th 2010, 10:09 AM
lvleph
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.
• Aug 24th 2010, 10:21 AM
yeKciM
Quote:

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 :D ) 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 $(2x^2)'= 2' \cdot x^2 + 2 \cdot (x^2)' = 0\cdot x^2 + 2\cdot 2x = 4x$ :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, 10:26 AM
lvleph
Quote:

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 :D ) 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.
• Aug 24th 2010, 10:43 AM
yeKciM
Quote:

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
• Aug 24th 2010, 10:46 AM
lvleph
No, it is not. Here is a good example. $\frac{d}{dx}\left[x\cdot f^2(x)\right] = x\cdot2f(x) \cdot f'(x) + f(x)$.
• Aug 24th 2010, 11:02 AM
yeKciM
Quote:

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 :D

sorry, but now i'm like confused :D

Edit: aaaah sorry I totally forgot about question up there... I'm very very sorry
• Aug 24th 2010, 11:11 AM
Quote:

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...