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

Answer 1

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

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})}}#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.