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differentiate the ffg wrt x: y = e^(-2x) . sin 5x y = 2cot (x^2 - 3) y = ln [x^2...

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donovanp | Student, Undergraduate | eNoter

Posted August 22, 2010 at 7:53 PM via web

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differentiate the ffg wrt x:

y = e^(-2x) . sin 5x

y = 2cot (x^2 - 3)

y = ln [x^2 .square root (1-sinx)]

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giorgiana1976 | College Teacher | Valedictorian

Posted August 22, 2010 at 7:57 PM (Answer #1)

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We'll differentiate the first composed function:

y = e^(-2x) . sin 5x

It is a product of 2 functions, so we'll differentiate according to the product rule:

(u*v)' = u'*v + u*v'

y' = [e^(-2x) . sin 5x]'

y' = [e^(-2x)]'*(sin 5x) + [e^(-2x)]*(sin 5x)'

y' = e^(-2x)*(-2x)'*(sin 5x) + [e^(-2x)]*(cos 5x)*(5x)'

y' = -2e^(-2x)*(sin 5x) + 5*[e^(-2x)]*(cos 5x)

We'll factorize and the final result will be:

y' = e^(-2x)*(-2sin 5x + 5 cos 5x)

 

We'll differentiate the second function y = 2cot (x^2 - 3)

y' = [2cot (x^2 - 3)]'

y' = -2(x^2 - 3)'/[sin (x^2 - 3)]^2

y' = -2*2x/[sin (x^2 - 3)]^2

y' = -4x/[sin (x^2 - 3)]^2

 

We'll differentiate the third given function

y = ln [x^2 *square root (1-sinx)]

y' = {ln [x^2 *square root (1-sinx)]}'

y' = [x^2 *square root (1-sinx)]'/x^2*square root (1-sinx)]

Since the numerator is a product, we'll apply the product rule:

[x^2 *square root (1-sinx)]' = (x^2)'*sqrt(1-sinx)+(x^2)*(1-sinx)'

[x^2 *square root (1-sinx)]' = (2x)*sqrt(1-sinx)+(x^2)*(-cos x)

y' = [(2x)*sqrt(1-sinx)+(x^2)*(-cos x)]/x^2*square root (1-sinx)

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neela | High School Teacher | Valedictorian

Posted August 22, 2010 at 9:15 PM (Answer #2)

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To differentiate:

 y =e^(-2x)*sinx

We use  y = u(x)v(x) = u'(x) v(x) +u(x)v'(x)

u(x) = (e^(-2x)], u'(x) = e^(-2x) * (-2x)' = -2 e^(-2x)

v(x) =  sin5x . v'(x) = (sin5x)' = (cos5x)(5x)' = 5 cos5x

So y' =  [-2e^(-2x)]sin5x + e^(-2x)*5cos5x = e^(-2x){-2sin5x+5cosx}

2)

y = 2cot(x^2-3)

We use y' = [ f(u(x))]' = f'(u) * u'(x).

y' = {2cot(x^2-3) }' = 2 { 1/tan(x^2-3)}' *

= 2 { 1/tan^2 (x^2-3)} (tan(x^2-c)'

=2 sec^2(x^2-3)/tan^2 (x^2-3) } {x^2-3)}'

={2 /sin^2(x^2-3)}2x

=4x/sin^2 (x^2-3)

3)

y = ln [x^2 .square root (1-sinx)]

We use {f(u(x)} = f'(u)*u'(x).

y' = {ln [x^2*(1-x^2)]}'  = {lnx^2 +ln(1-x^2)^(1/2)}'

= {2lnx + (1/2) ln(1-x^2)}' as ln a^m = mloga.

= (2lnx)' +[(1/2)ln(1-x^2)]'

= 2(lnx)' + (1/2) ln(1-x^2) , as (k*f(x)]' = k*f'(x)

= 2/x +[(1/2)/(1-x)^2](1-x^2)'

= 2/x +[1/2(1-x^2)](-2)

=2/x  - 1/(1-x^2)

 

 

 

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted August 23, 2010 at 2:24 AM (Answer #3)

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y = e^(-2x) . sin 5x

Let y = u*v    such that:

u = e^-2x  ==>  u' = -2e^-2x

v = sin5x  ==> v' = 5cos5x

==> y'= u'v + uv'

           = (-2e^-2x)*sin5x + (e^-2x)*5cos5x

           = -2e^-2x sin5x =5e^-2x* cos5x

           = (e^-2x)*(-2sin5x + 5cos5x)

 

y = 2cot (x^2 -3)

Let y = 2cot(u)  such that:

u= x^2 -3  ==> u' = 2x

==> y' = y'(u) * u'(x)

             = (2x)*-2/(x^2-3)^2 =

           = -4x/(x^2-3)^2

 

y = ln [x^2 .square root (1-sinx)]

Let u = x^2 * sqrt(1-sinx)

==> u' = 2x(cosx)

=>

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