Fine the derivative of the function:

1. g(x)=(1+4x)^5 (3+x-x^2)^8

2. y(x)=2^(sin(pie)x)

3. (1+cos^2x)^6

i dont get this **** and my test is on thursday.

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- October 17th 2006, 05:05 PMjephChain rule stuff
Fine the derivative of the function:

1. g(x)=(1+4x)^5 (3+x-x^2)^8

2. y(x)=2^(sin(pie)x)

3. (1+cos^2x)^6

i dont get this **** and my test is on thursday. - October 17th 2006, 05:57 PMThePerfectHacker
- October 17th 2006, 06:02 PMThePerfectHacker
We can express 2^x as e^(ln 2 * x)

Thus,

2^[sin (pi x)]=e^[ln 2 *sin (pi x)]

Now, the chain rule says that,

e^{f(x)} then the derivative is, f'(x)e^{f(x)}

That is take derivate of exponent and bring it down.

In this case the exponent is,

ln 2* sin(pi x) where ln 2 is a constant function,

The derivative is after using chain rule,

pi*ln 2*cos(pi x)

Thus, the answer is,

(pi*ln 2*cos(pi x))e^[ ln 2*sin (pi x)]

But, the 'e' exponent simplifies to (look at beginning of post),

(pi*ln 2*cos(pi x))2^[sin (pi x)] - October 17th 2006, 06:07 PMThePerfectHacker
You use a double chain rule here.

The inner function is,

1+[cos x]^2

The derivative of this function is,

-2 sin x cos x=-sin 2x (double angle)

Thus, derivative of

(1+cos^2 x)^6 is,

Inner function derivative which we found to be,

-sin 2x

Times outer function derivative,

6(1+cos^2 x)^5

Multiply them out,

-6*sin 2x(1+cos^2 x)^5 - October 17th 2006, 08:43 PMjeph
sorry about using that word, i didnt think it would have been considered inappropriate =/

can you show me how

1+[cos x]^2 = -2 sin x cos x

this whole time i tohught i wasnt getting the chain rule, but this is the part that got me.

the answer given was -12cosxsinx(1+cos^2x)^5 - October 18th 2006, 07:00 AMtopsquark
New questions should be posted in new threads!!

Are you asking how to take the derivative of 1 + [cos(x)]^2? Because

1+[cos x]^2 is most certainly NOT -2 sin x cos x. (For starters the LHS is always positive.)

d/dx{1 + [cos(x)]^2} = 0 + d/dx{[cos(x)]^2}

Your "outer function" is y^2 where y = cos(x) and your "inner function" is cos(x), so take the derivative of y^2 (2*y) and multiply that by the derivative of cos(x): ( [f(g(x))]' = f'(g(x))*g'(x) )

d/dx{[cos(x)]^2} = 2*cos(x)* -sin(x) = -2sin(x)cos(x)

-Dan - October 18th 2006, 08:29 AMThePerfectHacker
- October 18th 2006, 09:27 AMtopsquark