Evaluate and simplify y'.

y = ln |sec 3x...

Question

Evaluate and simplify y'.

y = ln |sec 3x|

Accepted Answer

Welcome back, everyone. Find the derivative of Y equals Ln of the absolute value of CC and 5 X with respect to X. We are given four answer choices A says -5 cotenent 5 X, B, 5 co tension 5X, C-5 5 X, and D 5 tangent 5 X. So first of all, we have to understand that. We're given the absolute value function, so we have to consider two cases because this is a piecewise function. Now first of all, let's assume that coequant of 5 X is greater than 0, right? The domain of a log essentially wants us to make sure that the function within the log is positive, right? The absolute value makes any value positive. Of course we want to exclude zeros. But first of all, let's just consider cosequence of 5 X being greater than 0, and this means that for this case Y must be equal to LN of cosequant of 5 X, right? We can essentially drop the absolute value sign. And then for the second case, we're going to consider coc of 5 X being less than 0, which means that Y is going to be equal to LN of negative cos of 5 X based on the definition of a piecewise function. We're going to show that each derivative is basically the same and it doesn't matter that we have the absolute value sign. So let's begin with the first one. We want to differentiate LN. Of cosequence of 5 X and we want to apply the chain rule. So let's recall that the derivative of LN of X is 1 divided by X. So in this case our angle is cosequence of 5 X. So we get the 1 divided by cosequence of 5 X, and then we have to apply the chain rule. So we have to multiply that by the derivative of the inner function. Which is cosequant of 5 X. And now if we simplify, well, we're going to get 1 divided by cosequant of 5 X multiplied by. Now, the derivative of coequant, let's recall that from the basic derivatives table. The derivative of cosequant of X is negative cosequent cotangent. So we have to multiply by negative cosequant 5X cotangent 5 X. And then, because our angle is 5 X and not just X, it is not an independent variable, we have to apply the chain rule once again and multiply that by the derivative of 5 X. So now let's simplify. We're going to factor out the negative sign. The derivative of 5 X is simply 5 based on the power rule. We can also see that cosequence of 5 X can be canceled out, so we will not have a denominator, we can just write negative 5 in front. And then we end up with cotangent. 5 X. Right. And now let's consider the second case. If we take the derivative of. Allan of negative cosequant of 5 X. We're simply going to repeat those steps, but notice that now we're going to get 1 divided by negative cosequant of 5 X, which is then multiplied by the derivative of negative cosequence of 5 X, which is going to have a positive sign, so cosequant of 5 X cotangent of 5 X, which is then multiplied by the derivative of. The angle 5 x right, and that's simply 5. So notice that we still get a negative sign. We are basically flipping the signs now our denominator has a negative sign, and we are multiplying by cossek and cotangent with a positive sign and overall. The product is still going to be negative, so we're going to get -5 cotangent 5X. So, it doesn't matter which part of the piece wise function we evaluate, we're still getting the same derivative, right? And overall, we can see that for any X within the domain of this function, The derivative is -5 cotangent 5X which corresponds to the answer choice A. Thank you for watching.

Evaluate and simplify y'.

y = ln |sec 3x|

Key Concepts

Derivative of Natural Logarithm

Chain Rule

Trigonometric Derivatives

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