Question 223602
To find dy/dx we will need to understand:<ul><li>Derivatives of a sum of functions</li><li>Derivatives of a product of functions</li><li>The Chain Rule</li><li>Implicit differentiation</li></ul>
I hope you understand most of these because I am not going to go into great detail.<br>
We'll start by<ul><li>Presuming that y is a function of x</li><li>Breaking the rest of the equation into "bite-size" functions of x. (By "bite-size" I mean functions for which we can easily find the derivative.)</li></ul>
So let's define u(x) = 1 and v(x) = {{{x^2e^y}}}. This allows us to rewrite the equation as:
y = u + v
And the derivative is easy for sums of functions:
Equation y' = u' + v'
u' is easy but v' is not. So we will break v into smaller pieces. Let's define {{{p(x) = x^2}}} and {{{q(x) = e^y}}}. This makes v = p*q. And the derivative of this product of functions:
v' = p*q' + q*p'
By the Chain rule,
q' = {{{e^y*y'}}}
and p' is an easy derivative:
p' = 2x
so by substituting these into v' = p*q' + q*p' we get:
v' = {{{x^2*e^y}}}*y' + {{{e^y*2x}}}
Now by substituting u' (which is 0) and v' (above) into y' = u' + v' we get:
y' = {{{0 + x^2*e^y}}}*y' + {{{e^y*2x}}}
which simplifies to:
y' = {{{x^2*e^y}}}*y' + {{{2xe^y}}}
Now all we need to do is solve for y'. Start by subtracting {{{x^2*e^y}}}*y' from both sides. This will gather the y' terms on the left side:
y' - {{{x^2*e^y}}}*y' = {{{2xe^y}}}
Next we'll factor out y' on the left side:
y'*{{{(1 - x^2*e^y)}}} = {{{2xe^y}}}
And finally divide both sides by {{{(1 - x^2*e^y)}}}:
y' = dy/dx = {{{(2xe^y)/(1 - x^2*e^y)}}}