pugola12 said:Homework Statement
Find dy/dx of y=(cosx)^x
Homework Equations
d/dx of a^u=lna*a^u*u'
The Attempt at a Solution
I thougt I just had to follow the form shown above, and this is what I got.
y=(cosx)^x
dy/dx=ln(cosx)*(cosx)^x*1
dy/dx=ln(cosx)*(cosx)^x
However, the actual answer is (cosx)^x*(ln(cosx)-xtanx)
I don't understand where this comes from at all. Thanks for your input.
The derivative of y=(cosx)^x is given by the expression dy/dx = (cosx)^x * (-sinx * ln(cosx) + (cosx)^(x-1) * sinx).
To find the derivative of a function with a power of a trigonometric function, you can use the chain rule and logarithmic differentiation. First, rewrite the function as y=e^(ln(cosx)^x) and then use the product rule to find the derivative.
The derivative of (cosx)^x is not simply x(cosx)^(x-1) because the power rule only applies to functions with constant exponents. In this case, the exponent is a variable, x, so the chain rule must be used to find the derivative.
Yes, the derivative of y=(cosx)^x can be simplified further using algebraic manipulation. You can also use the fact that ln(cosx) can be written as -ln(secx) to simplify the expression.
Yes, the domain of y=(cosx)^x is all real numbers except x=0 and the range is limited to positive values, since the function is always positive for any input. Additionally, the range is limited by the domain of the natural logarithm, which is all positive real numbers.