Finding Derivatives: y' of (sinx)^x and More

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The discussion focuses on finding the intervals where the function y = |2 - x| is increasing or decreasing, clarifying that it decreases for x < 2 and increases for x > 2. Participants emphasize breaking down absolute value functions into piecewise definitions to facilitate differentiation. The conversation also touches on implicit differentiation for the equation tan(xy) = x² and the application of the chain rule for differentiating functions like (sin x)ⁿ. Additionally, there is a discussion about determining the behavior of the rational function x/(x² - 1) by identifying critical points and analyzing sign changes. Overall, the thread provides insights into derivative calculations and function behavior analysis.
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On what intervals is y= l 2-x l increasing or decreasing?? Wouldnt it be its always increasing cause of the absolute symbol, but that's not the answer.

Also find where x/(x^2) - 1 is increasing/decreasing??

Given tan(xy)= x^2, find y'

y' of (sinx)^x
 
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First question:

Break the absolute value function up into a piecewise function.

second question:

what is the first derivative of the function?

third question:

can you solve for y? Do you know the derivative of arctan?

Fourth question:

This is an application of the chain rule. Let U=Sin x thus dU=Cos X dx. Do you know what U^x is?

Good Luck.
 
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What does the graph of y= |x| look like? What does the graph of y= |2-x| look like?

Rather than solve tan(xy)= x2 for y, I would be inclined to use "implicit differentiation".
 
graph is a v, but i still don't know
 
On what intervals is y= l 2-x l increasing or decreasing?? Wouldnt it be its always increasing cause of the absolute symbol, but that's not the answer.

Check this out: let f(x)=|2-x|.

f(-4)=6
f(-3)=5
f(-2)=4
f(-1)=3
f(0)=2
f(1)=1
f(2)=0
f(3)=1
f(4)=2

The values decrease when x is less than 2 and increase when x is greater than 2. But in general you may not know if this is exactly the whole behavior of the function.

As per the other suggestion, realize that |x| is given by this piecewise definiton:

|x| is x if x >= 0 and is -x if x<0.

Therefore, |2-x| is 2-x if 2-x >=0 and is -(2-x) if 2-x<0. If you rearrange this a bit, you get this:

|2-x| is 2-x if x <=2 and is x-2 if x > 2.

Now differentiate that and see when it is positive and negative. Is it differentiable at x=2? Remember that when f'>0, f is increasing and when f'<0, f is decreasing. Same applies to this question:




Also find where x/(x^2) - 1 is increasing/decreasing??

I assume you mean x/(x^2 - 1) becase what you have just reduces to 1/x - 1. Either way, those derivatives are fairly straightforward and you use the derivatives as indicated above. The problem is that you have to see when a rational (fractional) algebraic expression is positive or negative. To do this follow these steps which should have been in your precalculus background:
1. determine the "cut points". These are all points where the rational expression MAY change sign. To do this, factor the numerator and denominator and locate all zeros (roots) of the num. and den. Those are places where the WHOLE rational expression MAY change sign.

2. Make a number line with ONLY the cut points on it, in order.

3. These cut points divide the line into regions. One textbook refers to a "persistence of sign" rule which states that the rational expression must have the same sign in those regions. Therefore, testing the rational expression at one point suffices to determine its sign for that WHOLE region. Put a plus or minus, as appropriate, above the number line in each region to indicate the sign of the first derivative. When +, f is increasing; when -, f is decreasing.

Given tan(xy)= x^2, find y'

Indeed use implicit differentiation. This is not much harder than the examples in your book I'm sure. The trick is to use the Chain Rule on tan(xy) and then the product rule because you have a product of x and y and then when you differentiate y use the Chain Rule again. (the derivative of y is y'.)

y' of (sinx)^x

A handy formula is this:
A^{B}=e^{B\ln A}.

Use the formula and then differentiate using the Chain Rule.

Cheers
 
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parwana said:
graph is a v, but i still don't know

Then I have to ask: do you know what "increasing" and "decreasing" mean? Knowing that the graph is a v should completely answer your question.
 
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