# Math Help - Derivative of Trignometric Equation

1. ## Derivative of Trignometric Equation

Find y' if $sin(2xy)+cos^2(xy)=x^2y^2+3y^2.$

I tried using Implicit Differentiation, but the answer I got was all complicated & I can't simplify it.

2. Originally Posted by StarlitxSunshine
Find y' if $sin(2xy)+cos^2(xy)=x^2y^2+3y^2.$

I tried using Implicit Differentiation, but the answer I got was all complicated & I can't simplify it.
The derivatives of the terms on the left side are as follows:

1st term: $cos(2xy)\frac{d}{dx}(2xy)=(2y+2xy')cos(2xy)$

2nd term: $\frac{d}{dx}cos^2(xy)=2cos(xy)\frac{d}{dx}cos(xy)$

See if you can find the derivative $\frac{d}{dx}cos(xy)$.

Is everything I've done so far consistent with what you've attempted?

3. Actually, I didn't find the derivative of the first term of the first term.

Sin (2xy) I simply put as Cos (2xy). Why are you taking the derivative of the inside function ?

4. Originally Posted by StarlitxSunshine
Actually, I didn't find the derivative of the first term of the first term.

Sin (2xy) I simply put as Cos (2xy). Why are you taking the derivative of the inside function ?
Chain Rule

5. Originally Posted by StarlitxSunshine
Actually, I didn't find the derivative of the first term of the first term.

Sin (2xy) I simply put as Cos (2xy). Why are you taking the derivative of the inside function ?
Chain rule, just as e^i*pi said. Always remember to use the chain rule whenever you see a function in the parenthesis. If it isn't just $x$, then you have to apply the chain rule. In this case, you have to apply a combination of the chain rule and implicit differentiation. So try to find the derivative $\frac{d}{dx}cos(xy)$

6. Oh. Whenever it isn't x...I think I get it.

Okay, that should be:

$
\frac{d}{dx}[cos(xy)] = -sin(xy)((x)(y')+(y)(1))
$

$
= -xy'sin(xy) -ysin(xy)
$

...Right ?

7. Originally Posted by StarlitxSunshine
Oh. Whenever it isn't x...I think I get it.

Okay, that should be:

$
\frac{d}{dx}[cos(xy)] = -sin(xy)((x)(y')+(y)(1))
$

$
= -xy'sin(xy) -ysin(xy)
$

...Right ?
Yup

8. So do you understand how I found the derviative of $cos^2(xy)$

???

That one is kind of confusing. Other than that, the right side of the equation is very basic. I'm sure you can figure that out. So now you just have to solve for $y'$.

9. Yup Yup, Sorry :33

I understood that part & was able to finish the question myself.

Thank you very much =)