# Implicit Differentiation of $\cos^3(y)$

I think I understand implicit differentiation outside of those problems involving trig functions but for some reason this problem is breaking my brain:

Assume that y is a function of x. Find $\frac{dy}{dx}$for $x^3+\cos^3(y)=5x$

My first move, using the chain rule, is:
$3x^2+3\cos^2(y)(-\sin(y))=5$

However the text says this should instead be:

$3x^2+3\cos^2(y)(-\sin(y))\frac{dy}{dx}=5$

I don't understand where or how the $\frac{dy}{dx}$ is being introduced.

To my mind, $\cos^3(y)$ = $(\cos(y))^3$ and so could be interpreted as:
$f(x)=x^3$
$g(x)=\cos(y)$
$F(x)=f(g(x))=\cos^3(y)$

Then, using the Chain Rule:
$F'(x)=f'(g(x))\cdot g'(x)=3\cos^2(y)(-\sin(y))$

But apparently that's wrong.

by the chaine rule we get $3x^2+3\cos(y(x))^2(-\sin(y(x))\cdot y'(x)=5$
• So we can read $cos^3(y)$ as $cos^3(y(x))$? – Nick Oct 18 '14 at 17:04
• yes since $y=y(x)$ – Dr. Sonnhard Graubner Oct 18 '14 at 17:05