# Thread: Implicit Differentiation of a Trignometric Function

1. ## Implicit Differentiation of a Trignometric Function

Use implicit differentiation to find the second derivative:

> $y+ sin y = x$

$dy/dx + cosy dy/dx = 1$

$dydx (1+cosy) = 1$

$dy/dx = 1/1+cosy$

$d^2y/dx^2 = /frac{1+cosy*0 - 1*siny*dy/dx}{(1+cosy)^2}$

$\frac{0-siny*dy/dx}{(1+cosy)^2}$

$\frac{siny}{(1+cosy)^3}$

What did I do wrong ?

2. Originally Posted by StarlitxSunshine
Use implicit differentiation to find the second derivative:

> $y+ sin y = x$

$dy/dx + cosy dy/dx = 1$

$dy/dx (1+cosy) = 1$

$dy/dx = 1/1+cosy$

Up to this place all's right, but now you have to differentiate the right side, and get $\frac{d^2y}{dx^2}=\frac{\sin x}{(1+\cos x)^2}$.
I can't see from where that 3 in the power of the denominator in what you call "the answer" comes...

Tonio

$d^2y/dx^2 = /frac{1+cosy*0 - 1*siny*dy/dx}{(1+cosy)^2}$

$\frac{0-siny*dy/dx}{(1+cosy)^2}$

$\frac{siny}{(1+cosy)^3}$

What did I do wrong ?

.

3. My textbook has answers to all the odd numbered questions. The denominator is definitely cubed.

& with the differentiation, I'm still getting -sin y not sin y.

4. Originally Posted by StarlitxSunshine
My textbook has answers to all the odd numbered questions. The denominator is definitely cubed.

& with the differentiation, I'm still getting -sin y not sin y.
It is a quotient derivative: $\left(\frac{u}{v}\right)'=\frac{u'v-v'u}{v^2}$, and since the derivative of cos x is -sin x , with the sign minus it must be plus...and squared in the denominator.
What book do you have?

Tonio

5. Oh! I see where I went wrong with the minus sign Oo Thank you =)