What is the derivative of sinx pwr cosx?

Answer 1

For the function:

y = sin(x)cos(x)

To find the derivative y', implicit differentiation must be used. To do this, both sides of the equation must be put into the argument of a natural logarithm:

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

by the properties of logarithms, this can also be expressed as:

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

deriving both sides of the equation yields:

(1/y)(y') = cos(x)(1/sin(x))(cos(x)) + -sin(x)ln(sin(x))

This derivative features two important things. The obvious thing is the product rule use to differentiate the right side of the equation. The left side of the equation brings into play the "implicit" differentiation part of this problem. The derivative of ln(y) is a chain rule. The derivative of just ln(y) is simply 1/y, but you must also multiply by the derivative of y, which is y'. so the total derivative of ln(y) is (1/y)(y').

solving for y' in the above, the following is found:

y' = y[(cos2(x)/sin(x)) - sin(x)ln(sin(x))] = y[cot(x)cos(x) - sin(x)ln(sin(x))]

y' = y[cot(x)cos(x) - sin(x)ln(sin(x))] = sin(x)cos(x)[cot(x)cos(x) - sin(x)ln(sin(x)) is the most succinct form of this derivative.