trig derivative help...

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February 21st 2010, 11:42 AM
maybnxtseasn

trig derivative help...

h(x) = sin2xcos2x
h'(x) = -4cos2xsin2x <--- doubt this is correct
--------------------
f(x) = cotx/sinx (unsure on how to do this....)

-thx
February 21st 2010, 11:58 AM
ione

Quote:

Originally Posted by maybnxtseasn

h(x) = sin2xcos2x
h'(x) = -4cos2xsin2x <--- doubt this is correct
--------------------
f(x) = cotx/sinx (unsure on how to do this....)

-thx

For the first one try using $<br /> 2\sin{2x}\cos{2x}=sin{4x}$

If you know the derivatives of cotx and sinx, then use the quotient rule on the second one
February 21st 2010, 12:02 PM
skeeter

Quote:

Originally Posted by maybnxtseasn

h(x) = sin2xcos2x
h'(x) = -4cos2xsin2x <--- doubt this is correct
--------------------
f(x) = cotx/sinx (unsure on how to do this....)

-thx

you need to use the product rule ...

$h(x) = \sin(2x) \cos(2x)$

$h'(x) = \sin(2x)[-2\sin(2x)] + \cos(2x)[2\cos(2x)]$

$h'(x) = 2[\cos^2(2x) - \sin^2(2x)] = 2\cos(4x)$

or use an identity ...

$h(x) = \frac{1}{2} \cdot 2\sin(2x)\cos(2x)$

$h(x) = \frac{1}{2} \sin(4x)$

$h'(x) = 2\cos(4x)$

note that $\frac{\cot{x}}{\sin{x}} = \cot{x}\csc{x}$

product rule again ...

$\cot{x}[-\csc{x}\cot{x}] + \csc{x}[-\csc^2{x}]$

$-\csc{x}[\cot^2{x} + \csc^2{x}]$

$-\csc{x}[2\csc^2{x} - 1]$

or ...

$\frac{\cot{x}}{\sin{x}} = \frac{\cos{x}}{\sin^2{x}}$

use the quotient and chain rules ...

$\frac{\sin^2{x}(-\sin{x}) - \cos{x}[2\sin{x}\cos{x}]}{\sin^4{x}}$

$\frac{-\sin{x}[\sin^2{x} + 2\cos^2{x}]}{\sin^4{x}}$

$-\frac{1+\cos^2{x}}{\sin^3{x}}$