Question

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

Answer 1

`-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

Answer 2

`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)})`