Hello ishant. Welcome to PF !ishant said:Homework Statement
Homework Equations
The Attempt at a Solution
y = tanx^cosx
lny = cosx(ln(tanx))
1/y(dy/dx) = (sec^2x * cosx)/tanx + ln(tanx)(-sinx)
at pi/4
1/y(dy/dx) = (1 * 1/sqrt(2))/1 + ln(1)(-sin(pi/4))
1/y(dy/dx) = 1/sqrt(2) + 0 ******* ln1 = 0
***
y = tanx^cosx => 1^(1/sqrt(2)) = 1 => 1/y = 1
***
dy/dx at pi/4 = 1/sqrt(2)
The answer is sqrt(2) NOT 1/sqrt(2)
What am I doing wrong?
SammyS said:Hello ishant. Welcome to PF !
What is sec2(π/4) ?
The formula for finding the differential of tan(x)^cos(x) is d(tan(x)^cos(x)) = cos(x) * tan(x) * (-sin(x)) * dx.
To calculate the value of the differential at pi/4, you would first plug in pi/4 for x in the formula d(tan(x)^cos(x)) = cos(x) * tan(x) * (-sin(x)) * dx. Then, you would solve for dx to find the numerical value of the differential at pi/4.
Finding the differential of tan(x)^cos(x) can help in understanding the behavior of this function and how it changes with respect to x. It can also be used in optimization problems to find the maximum or minimum values of the function.
Yes, the value of the differential at pi/4 can be negative. This would indicate that the function is decreasing at that point.
The differential of tan(x)^cos(x) can be used in physics and engineering applications to calculate rates of change or slopes of curves. It can also be used in economics to analyze the sensitivity of a function to changes in its variables.