7–28. Derivatives Evaluate the following derivatives.

Question

d/dt ((sin t)^{√t})

Accepted Answer

Welcome back, everyone. Evaluate the derivative of cosine oft raised to the power of T2d. For this problem, because our base is a function of T and our exponent is a function of T, we're going to use a logarithmic differentiation. So first of all, let's assume that Y equals our function, which is cosine of T, raises the power of T2. We're going to take the natural logarithm of both sides, so LN of Y is going to be equal to LN of cosine of T. Raise the power of T2, and then we can apply the rules of logarithms and specifically the power rule. So we can rewrite the expression as LN of Y equals T2d. That's our exponent which now becomes a variable in front of the log. So T2d multiplied by LN of cosine of T. From here we can differentiate both sides. So let's identify the derivative of the left hand side and the right hand side. On the left hand side, the derivative of LN of Y is going to be 1 divided by Y, multiplied by Y based on the chain rule because Y is a function of T, right? On the right hand side, we want to apply the product rule. So first of all, we take the derivative of T2, which is 2T multiplied by The second function, which is LN of cosine of T. And then, according to the product rule, we're adding T squared multiplied by the derivative of LN of cosine of T, which is going to be 1 divided by cosine of T. Multiplied by cosine of t derivative which is negative sine of T. This is where we once again need to apply the chain rule. And now let's go ahead and simplify it. So first of all, we have Y divided by Y. Equals 2 LN of cosine of T plus T2, and then we can replace plus T2 with actually minus T2 right because we have a negative sign later on and sine divided by cosine, while it is equal to tangent. Now let's go ahead and solve for Y. Y is going to be equal to Y multiplied by 2 T. Allen of cosine of T minus T squad tangent of T. We can also factor out T, so let's write that in front. So that'd be Y equals YT multiplied by a factor which is 2 LN of cosine of T minus T tangent T. And now the final step is to replace Y with. Cosine Of tea Raised the power of T squared. We can write T in front as well. So Y would be equal to T multiplied by cosine of T raises the power of T2. And the final factor would be 2 of cosine of t minus T. Tangent see. And that would be our final derivative for this problem. Thank you for watching.

7–28. Derivatives Evaluate the following derivatives.

d/dt ((sin t)^{√t})

Key Concepts

Chain Rule

Product Rule

Exponential Functions and Logarithmic Differentiation

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