How do you differentiate $f (x) = {cos e}^{4 x}$ $f(x)=cose^(4x)$ using the chain rule.?

Question

2292 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-07T10:34:29

$f'(x) = - 4 sin (e^(4x)) * e^(4x)$

Your chain can be defined as follows:

$f(x) = color(green)(cos) color(blue)(e^(color(red)(4x)))$

$color(white)(xxx)= cos e^(w(x)) color(white)(xxx) " where " w(x) = color(red)(4x)$

$color(white)(xxx)= cos v(w) color(white)(xxxiii) " where " v(w) = color(blue)(e^w)$

$color(white)(xxx)= u(v) color(white)(xxxxxxi) " where " u(v) = color(green)(cos(v))$

Thus, to compute the derivative, you need to build the derivatives of $w(x)$ , $v(w)$ and $u(v)$ :

$w(x) = 4x color(white)(xxx) => color(white)(xx) w'(x) = 4$

$v(w) = e^w color(white)(xxx) => color(white)(xx) v'(w) = e^w = e^(4x)$

$u(v) = cos v color(white)(xx) => color(white)(xx) u'(v) = - sin v = - sin(e^w) = - sin(e^(4x))$

Now, the only thing left to do is to multiply the three derivatives!

$f'(x) = u'(v) * v'(w) * w'(x)$
$color(white)(xxxx) = - sin (e^(4x)) * e^(4x) * 4$
$color(white)(xxxx) = - 4 sin (e^(4x)) * e^(4x)$

How do you differentiate f(x)=cose^(4x)f(x)=cose4xf(x)=cose^(4x) using the chain rule.?