Use implict diffeentiation to find dy/dx
x^2(cosy)-y^2=e^2x
$\displaystyle \frac{d}{dx} ( x^2(\cos(y) - y^2) = \frac{d}{dx} e^{2x} $
$\displaystyle \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!
$\displaystyle \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:
$\displaystyle \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:
$\displaystyle \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] $\displaystyle \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 $\displaystyle \frac{dy}{dx} = $.
well the first term on the LHS contains:
$\displaystyle \frac{d}{dx} \cos(y) $
The chain rule says that if you have a function $\displaystyle h(x) = f(g(x)) $, then $\displaystyle h'(x) = f'(g(x)) \times g'(x) $
In our case $\displaystyle g(x) = y $, and $\displaystyle f(g(x)) = \cos(g(x)) $
And hence $\displaystyle g'(x) = \frac{dy}{dx} $ and $\displaystyle f'(g(x)) = -\sin(g(x)) $
Hence
$\displaystyle \frac{d}{dx} \cos(y) = -\sin(y) \times \frac{dy}{dx} $
In the RHS, we have the same sort of idea.
$\displaystyle g(x) = e^x $ and $\displaystyle f(g(x)) = (g(x))^2 $
Hence:
$\displaystyle g'(x) = e^x $, and $\displaystyle f'(g(x)) = 2g(x) $
So the result is :
$\displaystyle frac{d}{dx} (e^x)^2 = e^x \times 2e^x = 2(e^x)^2 = 2e^{2x} $