1. ## Differentiation

Differentiate

$y=(cosx)^{3x}$ with respect to x.

Using the basic differentiation rules for elementary functions, how do I approach this? What can be defined as u?

Is it $(cosx)^u$ or $u^{3x}$ ?

2. Originally Posted by VonNemo19
Differentiate

$y=(cosx)^{3x}$ with respect to x.

Using the basic differentiation rules for elementary functions, how do I approach this? What can be defined as u?

Is it $(cosx)^u$ or $u^{3x}$ ?
Logarithmic differentiation will do here.

$ln y = ln {(cos {x})^{3x}}$

$ln y = 3x ln {cos{x}}$

Now implicitly differentiate, then isolate $\frac{dy}{dx}$

I hope this helps.

3. That's what I thought, but I've never applied logs to trig functions before, so i was hesitant. Thanx.

Difeerentiating

$\frac{d}{dx}(lny)=\frac{d}{dx}(3xlncosx)$

$\frac{y'}{y}=3x\frac{-sinx}{cosx}+3lncosx$

y=3xlncosx

$y'=9xlncosx(\frac{-xsinx}{cosx}+lncosx)$

Is this correct?

4. Originally Posted by VonNemo19
That's what I thought, but I've never applied logs to trig functions before, so i was hesitant. Thanx.

Difeerentiating

$\frac{d}{dx}(lny)=\frac{d}{dx}(3xlncosx)$

$\frac{y'}{y}=3x\frac{-sinx}{cosx}+3lncosx$

y=3xlncosx

$y'=9xlncosx(\frac{-xsinx}{cosx}+lncosx)$

Is this correct?
No. I have marked in red what is uncorrect.

5. Originally Posted by running-gag
No. I have marked in red what is uncorrect.

Duh! Unbelievable. Thanx

6. Let u = cos(x) and v = 3x

then differentiate $y= u^{v}$ by using the chain rule for functions of more than one variable

$\frac {dy}{dx} = \frac {\partial y}{\partial u} \ \frac {du}{dx} + \frac {\partial y}{\partial v} \ \frac {dv}{dx}$