(Cosx)^y=x^siny find dy/dx at point pi/(4,0)
(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))
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...
To find dy/dx at the point (pi/4, 0), we can use implicit differentiation. First, let's differentiate both sides of the equation.
Differentiating (Cosx)^y = x^sin(y) implicitly with respect to x:
Using the chain rule, we get:
(d/dx) ((Cosx)^y) = (d/dx) (x^sin(y))
To differentiate (Cosx)^y, we can use logarithmic differentiation. Taking the natural logarithm of both sides, we get:
ln((Cosx)^y) = ln(x^sin(y))
y * ln(Cosx) = sin(y) * ln(x)
Now, let's differentiate both sides with respect to x:
(d/dx) (y * ln(Cosx)) = (d/dx) (sin(y) * ln(x))
Using the product rule, we can differentiate the left side:
(1 * ln(Cosx) + y * (-sinx / Cosx) * (-1/Cosx)) = (d/dx) (sin(y) * ln(x))
Simplifying the left side:
ln(Cosx) - y * sinx / Cos^2(x) = (d/dx) (sin(y) * ln(x))
Next, let's differentiate the right side:
(sin(y) * (1/x) + cos(y) * ln(x))
Setting the left side equal to the right side, we have:
ln(Cosx) - y * sinx / Cos^2(x) = sin(y) / x + cos(y) * ln(x)
Now, we can substitute the given point (pi/4, 0) into our equation:
ln(Cos(pi/4)) - 0 * sin(pi/4) / Cos^2(pi/4) = sin(0) / (pi/4) + cos(0) * ln(pi/4)
ln(1/√2) - 0 * 1 / (√2)^2 = 0 / (pi/4) + 1 * ln(pi/4)
ln(1/√2) = 0 + ln(pi/4)
ln(1/√2) = ln(pi/4)
Since both sides of the equation are equal, we can conclude that the point (pi/4, 0) satisfies the given equation.
Therefore, at the point (pi/4, 0), dy/dx = sin(y) / x + cos(y) * ln(x) is satisfied. However, since we were not given the value of y at the point (pi/4, 0), we cannot determine the exact value of dy/dx at that particular point without further information.