Question

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

Answer 1

[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})}}`