# Math Help - Implicit Differentiation & Indefinite Integral :(

1. ## Implicit Differentiation & Indefinite Integral :(

Help guys - I have loads of these to do and I can't get my head to work. I would really appreciate some detailed help so I can do the rest without exploing

sin(2x+3y)=3x^3y^2+4
imp. dif. to get dy/dx=f(x,y)

and

find the indefinite integral of g(x)=1 / (x^2 - 25)

for (-5 < x < 5 )

2. The trick for the first problem is to look at both sides of the equation as functions of $x$ and differentiate them, using the Chain Rule for $y$.

This gives us

$\frac{d}{dx}\sin(2x + 3y) = \frac{d}{dx}(3x^3y^2 + 4)$
$\cos(2x + 3y)\cdot 3\frac{dy}{dx} = 9x^2y^2 + 3x^3\cdot 2y\frac{dy}{dx}$
$\frac{dy}{dx}(\cos(2x + 3y)\cdot3 - 3x^3\cdot2y) = 9x^2y^2$
$\frac{dy}{dx} = 3x^2y^2\cdot(\cos(2x + 3y) - 2x^3y)^{-1}.$

For the second problem, it is useful to ask what constants $A$ and $B$ could make the equation

$\frac{1}{(x + 5)(x - 5)} = \frac{A}{x + 5} + \frac{B}{x - 5}$

true. This way, the integrands are simplified. In finding $A$ and $B$, we use the fact that the equation must hold for all $x$, not just one value.

3. Originally Posted by Scott H
The trick for the first problem is to look at both sides of the equation as functions of $x$ and differentiate them, using the Chain Rule for $y$.

This gives us

$\frac{d}{dx}\sin(2x + 3y) = \frac{d}{dx}(3x^3y^2 + 4)$
$\cos(2x + 3y)\cdot 3\frac{dy}{dx} = 9x^2y^2 + 3x^3\cdot 2y\frac{dy}{dx}$
$\frac{dy}{dx}(\cos(2x + 3y)\cdot3 - 3x^3\cdot2y) = 9x^2y^2$
$\frac{dy}{dx} = 3x^2y^2\cdot(\cos(2x + 3y) - 2x^3y)^{-1}.$

For the second problem, it is useful to ask what constants $A$ and $B$ could make the equation

$\frac{1}{(x + 5)(x - 5)} = \frac{A}{x + 5} + \frac{B}{x - 5}$

true. This way, the integrands are simplified. In finding $A$ and $B$, we use the fact that the equation must hold for all $x$, not just one value.
I'm not sure exactly what the question means by getting the answer into dy/dx =f(x,y) Does it just mean that it is a function of x and y in one expression?

Am I being very dense?!

Thanks so much for your help; I shall try and go through it at once!

4. That means it's okay to have $y$ on the other side of the equation.

5. Originally Posted by Scott H
That means it's okay to have $y$ on the other side of the equation.
so your final line of working is in the right format?

Thanks man

<tries to apply this to loads more!>