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

Question

𝔂 = 2 tan² x - sec² x

Accepted Answer

Welcome back everyone. Calculate the derivative of the function G of T equals 5 squared T plus cos 2 T. Let's write down the derivative we want to identify G T, which is going to be equal to 5 multiplied by the derivative of cotention squared of T. Let's add the derivative sign and then plus the derivative of cocC 2 C. What we can see is that we need 2 derivatives, and those are two composite functions. Let's begin with the first one. So what is the derivative of cotangent squared of T? While it is a composite function, we will need to apply the chain rule which says that we have to recognize that the outer function is the square and the inner function is the cotangent. According to the chain rule, first of all, we are applying the power rule for the square, so we're bringing down the exponent, which gives us 2 as a coefficient. We are decreasing the original exponent by 1 unit, so 2 minus 1 gives us 1, meaning we end up with cotangent, and then, according to the chain rule, we are multiplying by the derivative of cotangent of T. Now we're going to get two cotangent of T. What is the derivative of cotangent of T? Based on the table of basic derivatives, we know that it is equal to negative cosequant squad. Of tea. And now we can write the complete derivative here, right? So we're going to get -2. Cottingent of tan squared of t. Now let's continue and let's identify the derivative of coc2 of T. We're going to apply the same rule. First of all, the power rule, which gives us two cosequant of T multiplied by the derivative of the inner function. In this case it is the derivative of cosequant of T. So we're getting two cosequant of T. Multiplied by the derivative of cosequant of T. Now based on the table of basic derivatives, it is equal to negative cosequant of T cotangent of T. Overall, we're getting negative 2. Contingent of T. Coc can squad of T and now substituting these results into the original expression we are getting G T equals 5 multiplied by the derivative of quotation squared of T, which is -2. Cottingent of t cosquared of t. Plus, our next derivative, which is negative to cotangent of T co2d of T. Now let's multiply. We get negative 10. Cotingent of Tc can squared of t. Plus minus this means we're subtracting 2. Contingent of tea. Cozy and squared of tea. All that we have to do is combine the like terms. -10 minus 2 gives us -12. Cottingent of Tc can squad of t. Which is our final answer to this problem. Thank you for watching.

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

𝔂 = 2 tan² x - sec² x

Key Concepts

Derivative

Trigonometric Functions

Chain Rule

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