7–28. Derivatives Evaluate the following derivatives.

Question

d/dy (y^{sin y})

Accepted Answer

Welcome back, everyone. Evaluate the derivative of X raises the power of cosine x. For this problem, we're going to use a method called logarithmic differentiation. The first step is to assigny to our function. In other words, Y equals X, erases the power of cosine x. And then we can take the natural logarithm of both sides, so that LN of Y is going to be equal to LN of X, raises the power of cosine X. According to the properties of logarithms we can apply the power rule so that the right hand side becomes. Cosine of x multiplied by ln of X and that we can see. The derivative of each side, right? On the left hand side, the derivative would be one divided by Y multiplied by Y based on the chain rule. On the right hand side we can apply the product rule. So we take the derivative of cosine of X, multiply, I'm sorry, multiply by L of X, and add cosine of X multiplied by the derivative of LN X. So now let's identify each derivative. On the left hand side, we have Y divided by Y equals. On the right hand side, the derivative of cosine is negative sine of X. We multiply that by Ellen of X. And then we're adding cosine of X, which is multiplied by the derivative of LN of X, which is 1 divided by X. From here, we can express Y, so Y is going to be equal to Y multiplied by. Cosine of x divided by X minus. SinovX LN X. So we're multiplying both sides by a Y, right? Now, what we can do from here is basically express Y in terms of X, raise the power of cosine X. So Y equals. X raises the power of cosine x. Multiplied by Now, what we're going to do is simply find the common denominator to avoid the fractional term. So, in paris we get cosine of X minus sine XL and X must be multiplied by X, so we're subtracting X, sine x, LN X. And all of this is divided by X. Now, the reason why this is useful is because we're gonna actually divide X cosine X by X, right? So, Y. is going to be equal to X, cosine of X, divided by X, right? According to the properties of exponents, we can just write X to the power of cosine of X minus 1. Because when we have the same base in two different exponents. Upon division, we can just subtract those exponents, so cosine of X minus 1 because X in the denominator has an exponent of 1. So the final answer would be Y. Our derivative is equal to x the power of cosine x minus 1. Multiplied by the numerator, cosine of x minus X X X LNX. That'll be our final answer for this problem, and thank you for watching.

7–28. Derivatives Evaluate the following derivatives.

d/dy (y^{sin y})

Key Concepts

Derivatives

Chain Rule

Exponential Functions

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