How do you differentiate #f(x)=e^x*sin3x# using the product rule?

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Answer 1 · 2018-05-12T11:56:00

The answer is #=e^x(sin(3x)+3cos(3x))#

#(uv)'=u'v+uv'#

Here,

#f(x)=e^xsin(3x)#

and

#u=e^x#, #=>#, #u'=e^x#

#v=sin(3x)#, #=>#, #v'=3cos(3x)#

Therefore,

#f'(x)=e^xsin(3x)+3e^xcos(3x)#

#=e^x(sin(3x)+3cos(3x))#

Answer 2 · 2018-05-12T12:03:27

#e^x(sin 3x + 3cos 3x)#

We know,

If #f(x) = g(x) * h(x)#,

Then, #f'(x) = d/dxf(x) = d/dx(g(x) * h(x)) = d/dxg(x) * h(x) + g(x) * d/dxh(x)#.

This is the Product Rule for Differentiation.

Now,

#f'(x)#

#= d/dxf(x)#

# = d/dx(e^x * sin 3x)#

# = d/dxe^x * sin 3x + d/dx(sin 3x) * e^x#

# = e^x sin 3x + e^x cos 3x * d/dx(3x)# [Chain Rule]

#= e^x(sin 3x + 3cos 3x)#

Hope this helps.