(Cosx)^y=x^siny find dy/dx at point pi/(4,0)

Answer 1

(cosx)^y = x^siny
As you recall (or can work out yourself with little effort) the derivative of u^v where u and v are both functions of x, is just a generalization of the power rule and the exponential rule:

y = u^v
y' = v u^(v-1) u' + ln(u) u^v v'

Note that if u or v is a constant, you get the more familiar power and exponential rules.

(cosx)^y = x^siny
y (cosx)^(y-1) (-sinx) + ln(cos x) (cosx)^y y' = siny x^(siny-1) + lnx x^(siny) cosy y'

So, at (π/4,0)

0 + 0 = 1/√2 (π/4)^(1/√2 - 1) + ln(π/4) (π/4)^(1/√2) y'

y' = (2√2) / (π ln(4/π))

Answer 2

or, we might go at it using logs:

y ln(cos(x)) = sin(y) ln(x)

Now all we have to worry about is the chain rule and product rule:

ln(cos(x)) y' - y tan(x) = cos(y) ln(x) y' + sin(y)/x

y' =

y tan(x) + sin(y)/x
------------------------------
ln(cos(x)) - cos(y)ln(x)

= 0 at the indicated point.

Hmm. Clearly one of my answers is wrong.

In fact, the graph seems to indicate a vertical tangent. Now you have your work cut out for you...