Find the derivatives of the functions in Exercises 1–42.

Question

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𝓻 = √2θ sinθ

Accepted Answer

Welcome back everyone. Calculate the derivative of the function Y equals 4 T tangent of T. Raise the power of 14. So essentially what we're going to do is differentiate the original function Y, and this means that we're taking the derivative of 4 T. Multiplied by tangent of T raise the power of 1/4. First of all, we have to understand that we have an inner function and the outer function. So according to the chain rule, we first of all have to differentiate our inner function, right? With respect to 1/4, and it is an exponent, and this means that we're applying the power rule. So according to the power rule, we are taking 14th and bringing it down so it becomes a coefficient. 14th is then multiplied by word T multiplied by tangent of T. Raise the power of 1/4 minus 1. So here we have applied the power rule, and because we have an inner function, according to the chain rule, we have to multiply by the derivative of 4 T tangent of T. So now let's focus on the remaining derivative, and here we can apply the product rule. We have 4 T multiplied by tangent of T. According to the Product rule we're taking the derivative of the first function, so we're taking the derivative of word T, multiplying that derivative by the second function, which is agent of T. Then we are adding the first function which is 4 of T. And multiplying by the derivative of the second function or basically the derivative of tangent of t. Now according to the power rule, the derivative of T is 1, so we get 4. Tangent of tea. Plus the derivative of tangent acquiring the table of basic derivatives is c 2d, so we get plus or T22d of T. And then we're going to substitute this result into the original expression. This means that Y equals. 1/4 We get 4 T tangent of T raise the power of -34 multiplied by. We can factor out 4 and then we will have tangent. Of T plus. T 2 squared of T. We can notice that 1/4 and 4 those can be canceled out and now using the properties of exponents we can write this as a rational function because we have a negative exponent. So 4 tangent T goes into the denominator. Notice that we are raising it to the power of -34s originally, so now that exponent is going to be positive based on the reciprocal rule. Well then, we now have our denominator and our numerator is going to be tangent of T. Plus T Second squared of tea. Well then, we have identified the derivative for this problem, and thank you for watching.

Find the derivatives of the functions in Exercises 1–42.______𝓻 = √2θ sinθ

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Find the derivatives of the functions in Exercises 1–42.
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𝓻 = √2θ sinθ