1. ## Implict differentiation

Use implict diffeentiation to find dy/dx

x^2(cosy)-y^2=e^2x

2. Originally Posted by Emmeyh15@hotmail.com
Use implict diffeentiation to find dy/dx

x^2(cosy)-y^2=e^2x
$\frac{d}{dx} ( x^2(\cos(y) - y^2) = \frac{d}{dx} e^{2x}$

$\to \frac{d}{dx} ( x^2(\cos(y)) - \frac{d}{dx}(y^2) = \frac{d}{dx} e^{2x}$

The first term requies the product rule!

$\to x^2 \frac{d}{dx} (\cos(y)) + \cos(y)\frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = \frac{d}{dx} e^{2x}$

The RHS can be written:

$\to x^2 \frac{d}{dx} (\cos(y)) + \cos(y)\frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = \frac{d}{dx} (e^{x})^2$

And hence requires the chain rule:

$\to x^2 \frac{d}{dx} (\cos(y)) + \cos(y)\frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = 2e^x \times \frac{d}{dx} (2x)$

The first term also requires the chain rule:

[tex] $\to x^2 (\frac{d}{dy} (\cos(y)) \times \frac{d}{dx}(y)) + \cos(y)\frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = 2e^{x} \times \frac{d}{dx} (2x)$

Now all you have to do is carry out these relatively simple differentiations, and then change the equation so that you get $\frac{dy}{dx} =$.

3. im confused on how to apply the chain rule to that? i got the poblem right so far but i'm stuck at that point

4. Originally Posted by Emmeyh15@hotmail.com
im confused on how to apply the chain rule to that? i got the poblem right so far but i'm stuck at that point
well the first term on the LHS contains:

$\frac{d}{dx} \cos(y)$

The chain rule says that if you have a function $h(x) = f(g(x))$, then $h'(x) = f'(g(x)) \times g'(x)$

In our case $g(x) = y$, and $f(g(x)) = \cos(g(x))$

And hence $g'(x) = \frac{dy}{dx}$ and $f'(g(x)) = -\sin(g(x))$

Hence
$\frac{d}{dx} \cos(y) = -\sin(y) \times \frac{dy}{dx}$

In the RHS, we have the same sort of idea.

$g(x) = e^x$ and $f(g(x)) = (g(x))^2$

Hence:

$g'(x) = e^x$, and $f'(g(x)) = 2g(x)$

So the result is :

$frac{d}{dx} (e^x)^2 = e^x \times 2e^x = 2(e^x)^2 = 2e^{2x}$