differentiate w.r.t 'x'

Question

Cos(logx + e^x)

Answer 1

Hooray, a calculus question! Don't let it get to you, Kanwalpreet -- it's just as scared of you as you are of it!

The key is to know the Chain Rule version of the derivative formulas, in this case the Cosine rule:

d/dx [CosU] = -SinU * U'

You already have the Cosine, so we need only identify the U and then do fun things to it. In this case, the U must be everything inside the Cosine:

U = Log(x) + e^X

The fun of the Chain rule is that you now get to employ additional derivative rules to finish off this thing:

d/dx [Log V] = V' / V                            d/dx [e^W] = e^W * W'

Please note that I only used the letters V and W to avoid confusion with the U. In reality, we normally write "U" for every rule. Using other dummy variables is a perfectly legal trick to keep things tidy. Now, we must identify our V and W so we can do even more fun things to them:

V = X                                                       W = x

V' = 1                                                       W' = 1

Wow, that was exhausting! Let's start plugging things back in now:

d/dx [LogX] = 1/X                           d/dx[e^X] = e^X * 1

Finally, this means we can complete the original question:

d/dx [Cos (LogX + e^X] = -Sin(LogX + e^X] * (1/X + e^X]

See? We just derived the outer function, in this case the trig, then everything inside of it as well and, if there had been anything more complex than "X", derive them as well...