In Exercises 57–70, use logarithmic differentiation to find th...

Question

In Exercises 57–70, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.

61. y = √(θ + 3) sin θ

Accepted Answer

Welcome back everyone. Use logarithmic differentiation to find D Y divided by DX for Y equals square root of X2 + 4 multiplied by tangent of X. For this problem, as the problem suggests, we're going to use logarithmic differentiation, which means that we're going to take LN of both sides of this equation. So we take LN of Y, which is equal to LN of square root of X2 + 4 multiplied by tangent of X. What we can do is apply the properties of logarithms to show that LN of Y is equal to LN square root of X2 + 4, plus, using the product rule. Allen of tangent of X, right? Now, additionally, we can apply the power rule by rewriting x2 + 4 under the radical with x2 + 4 to the power of 1/2, and this allows us to bring down the exponents so we can write 1/2 m. Of x2 + 4. And then we're going to take the derivative of each side. So LN of Y derivative is 1/2 LN X2 + 4 derivative plus LN of tangent X derivative. Let's find each derivative. Starting with LN of Y derivative, we have to apply the chain rule because Y is a function of X, it is. Basically a function of an independent variable. So we differentiate LN of Y with respect to Y, which gives us one divided by Y. And according to The chain rule we're multiplying by the derivative of Y. Now for LN of X2 + 4 derivative, once again we're going to apply the chain rule which gives us 1 divided by x2 + 4 multiplied by x2 + 4 derivative, and this is equal to 2 x divided by X2 + 4. Then we have a land of the ancient X derivative. Once again applying the chain rule we get 1 divided by tangent X. Multiplied by tangent x derivative. The derivative of tangent is c 2d x and we're dividing it by tangent x. So now we can substitute this into our equation to get. Y divided by y. Equals 1/2 multiplied by 2 X Divided by X2 + 4. Plus Second squared x divided by tangent X. Notice that we can simplify slightly, we can cancel out too, right? And we can basically show that this is equal to X. Divided by x2 + 4. Plus 2 squared x divided by tangent X. And then we can solve for Y by multiplying both sides of this equation by Y. And knowing that y is equal to square root of. X2 + 4 multiplied by tangent X from the description of the problem, right. We can multiply this function by x. Divided by x2 + 4. Plus 2 squared x. Divided by 1 X. What we can now do is distribute. Let's go ahead and do that. So Y is equal to. X Multiplied by a square root of X2 + 4 tangent X. Divided by X2 + 4 + Second squared X multiplied by square root of. X2 + 4 tangent x divided by tangent X. Notice that we can simplify slightly. For the first term, we can cross out square root because in the denominator we have the same term. Raise the power of one, so we simply get square root of. X2 + 4 in the denominator. And for the second term, we can cross out tents. And finally, we're going to use the common denominator to rewrite our answers, so that Y, which is our derivative. It is going to have a common denominator of square root of X2 + 4. We're going to rewrite the first. Numerator, which is X tangent X. And we're adding c square x multiplied by square root of X2 + 4 multiplied by itself because we do not have any denominator and we're using a common denominator of square root of X2 + 4. So we're basically squaring the radical. To get X2 + 4. We can now rearrange our terms so that we're going to get X. Tangent X plus. X2 + 4 multiplied by c 2 X. Divided by square root of X2 + 4, which is our final answer for this problem. Thank you for watching.

In Exercises 57–70, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.61. y = √(θ + 3) sin θ

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In Exercises 57–70, use logarithmic differentiation to find the derivative of y with respect to the given independent variable.
61. y = √(θ + 3) sin θ