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Compute the derivative

0 votes

x^5*e^x*cos(3x)

asked Apr 19, 2013 in CALCULUS by mathgirl Apprentice

2 Answers

+1 vote
 
Best answer

Let y = x5 × ex × cos3x

Let u = x5 × ex

Apply derivative to each side with respective x

du / dx = ex(x5 + 5x4)

Substitute u = x5 × ex in the function y

y = u × cos3x

Apply derivative to each side with respective x

Recall :Derivative(UV) = UV' + VU'

dy / dx = u (-3sin3x) + du / dx (cos3x)

Substitute u = x5 × ex and du / dx = ex(x5 + 5x4) in the dy / dx

dy / dx = x5 × ex(-3sin3x) + ex(x5 + 5x4)cos3x

dy / dx = ex[-3x5sin3x + (x5 + 5x4)cos3x]

dy / dx = ex[(x5 + 5x4)cos3x - 3x5sin3x].

answered Apr 19, 2013 by diane Scholar
selected May 13, 2013 by mathgirl
0 votes

Given x^5*e^x*cos(3x)

Let y = x^5*e^x*cos(3x)

Differentiate with respect to x each side

Recall : Derivative (uvw)' = u*v*w' + u*v'*w + u'*v*w

From the above u = x^5 , v = e^x , w = cos(3x)

dy/dx = x^5*e^x*(cos(3x))' + x^5*(e^x)'*cos(3x) + (x^5)'*e^x*cos(3x)

dy/dx = x^5*e^x*(-3sin(3x)) + x^5*e^x*cos(3x) + (5x^5-1)e^x*cos(3x)

dy/dx = -3sin(3x)x^5e^x + x^5e^xcos(3x) + 5x^4e^xcos(3x)

dy/dx = e^x ( -3sin(3x)x^5 + x^5cos(3x) + 5x^4cos(3x) )

dy/dx = e^x ( -3sin(3x)x^5 + cos(3x)(x^5 + 5x^4) )

dy/dx = e^x ( cos(3x)(x^5 + 5x^4) - 3 x^5sin(3x) )

answered May 13, 2013 by jeevitha Novice

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