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# Find $y'$ and $y".$$y = \ln \mid \sec x \mid$

## $y^{\prime \prime}=\sec ^{2} x$

Derivatives

Differentiation

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### Video Transcript

So if this problem we want to do is take the derivative bond Ultimately this promise giving us practice with logarithmic functions. We know that logarithmic functions are used commonly in math applications, so it's useful that we can determine the rates of change it any human point. So in this case, what we're gonna have is the function of X being equal to the natural log of the absolute value of seeking X. We can put this in parentheses if we want, and that will give us the cross. Then when we take the derivative of this, what we're going tohave is one over seeking X. And then we also want to multiply that, um, by what we have in here because that's the chain role. So we'll have take an X tangent x um, because the seeking X will cancel with the seeking X. What we end up getting as a result is tangent X. So if they were to see after Prime of X, we see that it's equal Thio tangent X. If we were to just remove thes portions of the graph, we see that it matches up exactly. So that shows us that we did it correctly and that the derivative of the natural log of the absolute value of seeking X is Tangent X. And then we want Thio take the second derivative. So the second derivative of Tangent X is a trigger identity. And we know that to be seeking squared so we could graft f double climax right here. And we see that will be the same thing as seeking Squared X and as we see it does match up again.

California Baptist University

Derivatives

Differentiation

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