How do you differentiate #f(x)=sqrtsin(e^(4x))# using the chain rule.?

1 Answer
Feb 9, 2016

Answer:

[2e^(4x)cos(e(^4x))] / [sqrt sin(e^(4x)]

Explanation:

Applying chain rule
df(u)/dx= df/du .du/dx
let sin e^(4x) =u
d/du √u . d/dx (sin(e^(4x)))
we have,
d/du √u=1/(2√u)
and
d/dx (sin(e^(4x)))

Applying chain rule,
df(u)/dx= df/du .du/dx
let #e^{4x}=u#
#\frac{d}{du}# (sin(u) #\frac{d}{dx}#

#\frac{d}{du}# sin(u) = cos(u)
#\frac{d}{dx} (e^{4x}\)#
Applying chain rule,
df(u)/dx= df/du .du/dx
let 4x=u
solving it we get,
#e^{4x}#4
now, cos(u) #e^{4x}4#
u= #e^{4x}#
= cos(#e^{4x}#) #e^{4x}4#

Finally,
#\frac{1}{2\sqrt{u}}# cos(#e^{4x}#) #e^{4x}4#
Substitute u= sin( #(e^{4x})#
=#\frac{1}{2\sqrt{\sin (e^{4x})}}# cos( #(e^{4x})e^{4x}4#
simplifying we get,
#\frac{2e^{4x}\cos(e^{4x})}{\sqrt{\sin (e^{4x})}}#