Derivative of Trignometric Equation

**StarlitxSunshine** · November 8th 2009, 02:43 PM

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.

**adkinsjr** · November 8th 2009, 02:54 PM

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?

**StarlitxSunshine** · November 8th 2009, 02:56 PM

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 ?

**e^(i*pi)** · November 8th 2009, 03:04 PM

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

**adkinsjr** · November 8th 2009, 03:09 PM

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 $\frac{d}{dx}cos(xy)$

**StarlitxSunshine** · November 8th 2009, 03:18 PM

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 ?

**adkinsjr** · November 8th 2009, 03:22 PM

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

**adkinsjr** · November 8th 2009, 03:25 PM

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 $y'$ .

**StarlitxSunshine** · November 9th 2009, 03:41 PM

Yup Yup, Sorry :33

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

Thank you very much =)

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