In Exercises 13–24, find the derivative of y with respect to t...

Question

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.

19. y = (sech θ)(1-ln(sech θ))

Accepted Answer

Welcome back, everyone. Find D Y divided by the X for Y equals X multiplied by 2 minus LN of X. For this problem we want to find the Y divided by D X. Notice that we have a product of two functions, so we can apply the product rule. Let's remember that we are taking the derivative of the first function, so the derivative of sash of X, multiplied by the second function to minus LN of sesh X. And we are adding the first function, se of X. Multiplied by the derivative of the second one, so the derivative of 2 minus Ln of sax. From here, we want to find two derivatives. Let's begin with the derivative of X. Now, this is an elementary derivative. Its value is negative X X. And now for the remaining one, well, we can show that the derivative of 2 minus LN of X. The derivative of this expression is going to be equal to 0, right? The derivative of 2 is 0 because 2 is a constant. We can factor out the negative sign and focus on the derivative of LN. Of sesh X, right? Now, this is a composite function. First of all, we want to Differentiate natural log with respect to sesh. The derivative is a one divided by sesh X. Right? And according to the chain rule, we're multiplying by the derivative of the inner functions, so we're multiplying by the derivative of such X. We already know what the derivative of sesh X. Is So we can substitute that expression into the numerator. We're going to get negative, our first negative sign, followed by the next negative sign coming from such X derivative. In the numerator, we're going to add such X X. In the denominator, we have sesh X. So now we can cancel out two negative signs, we can cancel out sesh. And we're going to get change. So we're going to get 10X. And now let's use these derivatives. In the product rules so that the DY divided by DX is going to be X derivative, which is negative X. 0 X. Multiplied by 2 minus. Alan of Seche. X Plus Sesh X multiplied by 10 X. Now, what we're going to do is simplify, right? Let's factor out such X sta X because it's a repeating factor. So we're going to take out sesh X. 0 X. Which leaves us with one. Minus 2 minus negative LN of X, we're basically adding LN of X, right? And then 1 minus 2 is -1. So in pregnancies, we can just rearrange the terms so that we are starting with the positive one, that'd be LN of sesh. Eggs -1. So the final derivative is X X X multiplied by. Alanov X minus 1. Thank you for watching.

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.19. y = (sech θ)(1-ln(sech θ))

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In Exercises 13–24, find the derivative of y with respect to the appropriate variable.
19. y = (sech θ)(1-ln(sech θ))