How do you differentiate #f(x)=-5 xe^(x/cos x)# using the chain rule?

2 Answers
Jun 21, 2016

Answer:

# -5 e^{x/cos x} (1 + x (cos x + x sin x)/cos^2 x )#

Explanation:

you'll need the product rule first of course so we can simplify by letting #p = e^{x/cos x}#

then first we work the product rule so #(-5x p)' = -5 p -5x p'# Eqn A

and now for p', we know from the chain rule that #(e^{f(x)})' = f'(x) e^{f(x)}# where, here, #f(x) = x/ cosx#

and so
so # f'(x) = (x / cos x)' = (cos x + x sin x)/cos^2 x# from the quotient rule

which means that #p' = (e^{x/cos x})' = e^{x/cos x} (cos x + x sin x)/cos^2 x #

going back to Eqn A, and plugging this all back in, we then have

#(-5x e^{x/cos x})' = -5 e^{x/cos x} -5x e^{x/cos x} (cos x + x sin x)/cos^2 x#

#= -5 e^{x/cos x} (1 + x (cos x + x sin x)/cos^2 x )#

that could be tidied up maybe in some other way...

personally, i think it is helpful to look at logarithmic differentiation here

so we can instead say that #ln p = x/ cosx#

so #1/p p' = (x / cos x)' = (cos x + x sin x)/cos^2 x#

so # p' = p(x / cos x)' = e^{x/cos x} (cos x + x sin x)/cos^2 x# again using the quotient rule for the second bit of this.

another way of looking at it

Jun 21, 2016

Answer:

#frac{d}{dx}(-5xe^{frac{x}{cos (x)}})=-5(e^{frac{x}{cos (x)}}+frac{e^{frac{x}{cos (x)}}x(xsin (x)+cos (x))}{cos ^2(x)})#

Explanation:

#frac{d}{dx}(-5xe^{frac{x}{cos (x)}})#

Taking the constant out, #(acdot f)^'=a\cdot f^'#

#=-5frac{d}{dx}(xe^{frac{x}{cos (x)}})#

Applying the product rule, #\(f\cdot g\)^'=f^'\cdot g+f\cdot g^'#
#f=x,g=e^{frac{x}{cos (x)}}#

#=-5(frac{d}{dx}(x)e^{frac{x}{cos(x)}}+frac{d}{dx}(e^{frac{x}{cos(x)}})x)#

We know,
#frac{d}{dx}(x)=1#
#frac{d}{dx}(e^{frac{x}{cos (x)}})=frac{e^{\frac{x}{cos (x)}}(xsin (x)+cos (x))}{cos ^2(x)}#

[i.e applying chain rule; #frac{df(u)}{dx}=frac{df}{du}cdot frac{du}{dx}#

Let #frac{x}{cos (x)}=u#
#=frac{d}{du}(e^u)frac{d}{dx}(frac{x}{cos (x)})#
and, #frac{d}{du}(e^u)=e^u#
also, #frac{d}{dx}(frac{x}{cos (x)})=frac{cos (x)+xsin (x)}{cos ^2(x)}#

#=e^ufrac{cos (x)+xsin(x)}{cos ^2(x)}#

substituting back, #u=frac{x}{cos (x)}#
#=e^{frac{x}{cos (x)}}frac{cos(x)+xsin(x)}{cos ^2(x)}#

simplifying it,
#=frac{e^{frac{x}{cos (x)}}(xsin (x)+cos(x))}{cos ^2(x)}# ]

so, #=-5(1e^{frac{x}{cos (x)}}+frac{e^{frac{x}{cos(x)}}(xsin(x)+cos(x))}{cos ^2(x)}x)#

simplifying it,
#5(e^{frac{x}{cos (x)}}+frac{e^{frac{x}{cos (x)}}x(xsin (x)+cos (x))}{cos ^2(x)})#