In Exercises 25–36, find the derivative of y with respect to t...

Question

In Exercises 25–36, find the derivative of y with respect to the appropriate variable.

35. y = sinh⁻¹(tan x)

Accepted Answer

Welcome back everyone. Find dyy divided by dx if y equals inverse of sing of cotangent of x. For this problem, let's recognize that y is a composite function, so we basically want to apply the chain rule. We can show that the d y divided by d x. Is first of all the derivative of inverse of sin of cotangent of x with respect to the inner function so that would be cotangent of x and according to the general rule we are multiplying by the derivative of the inner function cotangent of x with respect to x. Now let's remember the derivative of inverse of sing. It is going to be 1 divided by. Square root of a one. Plus the square of the inner function, so 1 plus. Cotangent squared of x followed by the derivative of cotangent of x. Well, this is an elementary derivative and it is negative co sequin squared of x. Now all that we want to do is simplify, so we get 1 divided by a square root of let's remember trigonometric identities 1 plus cotangent squared of x is equal to co sequin squared of x. And we are multiplying by negative co sequence squared of x. We can write cosequent square of x in the numerator, and we can write negative in front. Now, let's go ahead and simplify that would be negative co sequin square of x divided by. The square root of co sequin squared of x is equal to the absolute value of cosequant of x. And from here we can use the property that co sequin 2 x can be written as the absolute value of co sequin square of x as well, right, because it's always non-negative. We're dividing by the absolute value of co sequent of x, and we can now combine those absolute values that would be negative. Absolute value of co sequin squared of x divided by cosequant of x. Which gives us co sequin square divided by cosequent reduces into co sequin, so that would be negative of the absolute value of co sequent of x, and that's our final answer for this problem. Thank you for watching.

In Exercises 25–36, find the derivative of y with respect to the appropriate variable.35. y = sinh⁻¹(tan x)

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In Exercises 25–36, find the derivative of y with respect to the appropriate variable.
35. y = sinh⁻¹(tan x)