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Solve the given problems.Find the linearization $L(x)$ of the function $f(x)=\sin (\cos x)$ for $a=\pi / 2$

Calculus 1 / AB

Chapter 27

Differentiation of Transcendental Functions

Section 1

Derivatives of the sine and Cosine Functions

Derivatives

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In mathematics, precalculu…

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In mathematics, a function…

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Find the linearization $L(…

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Find the linearization $ L…

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Okay. In this problem we are asked for the liberalisation Halifax. This function at the point A. So in order to do this, we need a slope and a coordinate. We'll start off with the coordinate so we know that X coordinate is paid over to In order to get the Y coordinate. We're just going to compute f of fire over to. So this is sign co sign a pie of the tube. Co sign a pie of the tube. We should know that that is zero. So this turns into sino zero, which we also know zero. So I coordinate that this linear ization needs to go through is pi over two common zero. Okay, Now comes the job of finding the slope. We'll call that m Okay, that's going to come from the derivative. So we need F prime of X first. So the derivative of sine is co sign. Okay. And then we chain onto this, the derivative of the inside the derivative in a function derivative of coastline we know is negative sign. Okay? We want the slope at a value. So pi over two. So we substitute pi over two in here. Okay, I don't need to worry too much about simplifying this down because we're gonna get a value out of here. Co signed a pi over two. We know zero sine of pi over two. We know that's one. Okay. So we're gonna end up with co sign zero times negative one coastline of zero. We know is one. So we got one times negative one. Just negative one. Okay, now we conform uh linear ization L. Of X. Using the point slope form. So y minus Y. One equals M X minus X. One. This is what we just found as I am. That's a slope. So why am I in a zero equals negative one. Explain this pie over to a little bit of simplification leads us to why you cause negative X was fired over two. And that is the equation of our linear ization of X.

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