# [SOLVED] Several questions

• Feb 4th 2010, 02:49 PM
Amberosia32
[SOLVED] Several questions
1) How do you integrate Sin^2 x or cos^2 x I know that the integral of sin x is -cos x and that the integral of cos x is sin x but I don't know what to do with the square.

2)How do you integrate e^5x I know that the integral of e^x is e^x so is it just e^5x?

3)How do you do this: Find the tangent line to: (x^4 + 6xy +y^2 = 29) I don't know what to do with the y's
• Feb 4th 2010, 04:30 PM
Sudharaka
Dear Amberosia32,

For the first and second problems you should use the chain rule. That is if y and z are functons of x, $\frac{dy}{dx}=\frac{dy}{dz}\times\frac{dz}{dx}$.

eg: $\frac{d(sin^{2}x)}{dx}=\frac{d(sinx)^2}{d(sinx)}\t imes{\frac{(sinx)}{dx}}=2sinxcosx=sin2x$

Similar method could be used to differentiate $cos^{2}x$ and $e^{5x}$

For a curve in two dimention(including a independent and dependent variable.) $\frac{dy}{dx}$ gives the tangent of the curve.
Therefore you could differentiate the equation, and afterwards subject $\frac{dy}{dx}$,

$x^{4} + 6xy +y^{2} = 29$

$4x^{3}+6x\frac{dy}{dx}+6y+2y\frac{dy}{dx}=0$

Subject $\frac{dy}{dx}$ and you would get,

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

Here I have used the Leibniz's theorem in addition to the chain rule,

That is if v and u are two functions of x, $\frac{d(vu)}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}$

• Feb 4th 2010, 04:35 PM
Amberosia32
Quote:

Originally Posted by Sudharaka
Dear Amberosia32,

For the first and second problems you should use the chain rule. That is if y and z are functons of x, $\frac{dy}{dx}=\frac{dy}{dz}\times\frac{dz}{dx}$.

eg: $\frac{d(sin^{2}x)}{dx}=\frac{d(sinx)^2}{d(sinx)}\t imes{\frac{(sinx)}{dx}}=2sinxcosx=sin2x$

Similar method could be used to differentiate $cos^{2}x$ and $e^{5x}$

For a curve in two dimention(including a independent and dependent variable.) $\frac{dy}{dx}$ gives the tangent of the curve.
Therefore you could differentiate the equation, and afterwards subject $\frac{dy}{dx}$,

$x^{4} + 6xy +y^{2} = 29$

$4x^{3}+6x\frac{dy}{dx}+6y+2y\frac{dy}{dx}=0$

Subject $\frac{dy}{dx}$ and you would get,

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

Here I have used the Leibniz's theorem in addition to the chain rule,

That is if v and u are two functions of x, $\frac{d(vu)}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}$

I don't understand how you got that 2sinxcosx=sin2x And in the other one were does the 6x or the 6y come from?
• Feb 4th 2010, 04:39 PM
pickslides
Quote:

Originally Posted by Amberosia32
I don't understand how you got that 2sinxcosx=sin2x

This is a standard trig identity, you should commit it to memory

Quote:

Originally Posted by Amberosia32
I And in the other one were does the 6x or the 6y come from?

Through implicit differentiation. Are you aware of this?
• Feb 4th 2010, 04:44 PM
Amberosia32
Quote:

Originally Posted by pickslides
This is a standard trig identity, you should commit it to memory

Through implicit differentiation. Are you aware of this?

never heard of it....My class is more of a figure it out your self setting. But thanks.