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# For each of the following function (i) give a definition like those in (2), (ii) sketch the graph, and (iii) find a formula similar to Equation 3.(a) $csch^{-1}$ (b) $sech^{-1}$ (c) $\coth^{-1}$

## $\frac{1}{2} \ln \left(\frac{x+1}{x-1}\right)$

Derivatives

Differentiation

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

okay. This problem, we're gonna use thes three formulas to get formulas for these three inverse, trig, hyperbolic trig functions. So we're going to start by setting Why, equal to the hyperbolic coast seeking inverse of X. Now take the hyperbolic coast seeking of both sides. So you get the hyperbolic coast. Second of y equals X. So, you know, the hyperbolic coast seeking is the same thing as one over the hyperbolic sign of why flip both these fractions upside down. So you get the hyperbolic sign of why is one over X Now take the hyperbolic sine inverse of both sides. All right, now we know what to do because we know a formula for the hyperbolic sine inverse. Everywhere you see X in the formula, put one over X. So this is equal to the natural log of one over X plus the square root of one over X squared plus one. All right, now we're going to find the formula for the hyperbolic seeking in verse what? We don't have to go through all those first steps because we can see that all you have to dio is put the reciprocal function and then the reciprocal um, argument in there. So you get the hyperbolic cosine inverse of one over X. Here's the formula. Fritz. Everywhere is he X Put one over X. So get the natural log of one over X plus the square root of one over X squared minus one. Okay. And then the hyperbolic co tangent in verse Why was the hyperbolic co tangent inverse of X is the reciprocal function so hyperbolic tangent in verse and the reciprocal argument one over X. So you get one half the natural log of 11 uh, of one plus one over X over one, minus one over X. Okay, you don't want to leave it like that with a four story fraction. So multiply everything in the fraction by X. It's like if you won half the natural log of X plus one over X minus one and there you formulas

Oklahoma State University

#### Topics

Derivatives

Differentiation

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp