Implicit Differentiation of a Trignometric Function

**StarlitxSunshine** · November 15th 2009, 03:14 PM

Use implicit differentiation to find the second derivative:

> $y+ sin y = x$

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

$dydx (1+cosy) = 1$

$dy/dx = 1/1+cosy$

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

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

But the answer is

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

What did I do wrong ?

**tonio** · November 15th 2009, 03:29 PM

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

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

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

But the answer is

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

What did I do wrong ?

.

**StarlitxSunshine** · November 15th 2009, 03:41 PM

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.

**tonio** · November 15th 2009, 04:11 PM

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

**StarlitxSunshine** · November 15th 2009, 04:27 PM

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

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