derivatives and areas: y = ln sqrt (1 - x^2)

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xc630

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Hello I would like some help with two problems

1) For absolute value of (x) < 1, the derivative of y= ln SQRT (1-x^2) is...

I have 1/ SQRT(1-x^2) for the regualkr derivative but I am not sure how "for absolute value of (x) < 1" would change the answer if it does.

2) The tangent line to the graph e^2-x at the point (1,e) intersects both coordinate axes. What is the area of the triangle formed by this tangent line and the coordinate axes?

I cannot use a calculator so can someone get me started
 
Re: derivative & areas

xc630 said:
1) For absolute value of (x) < 1, the derivative of y= ln SQRT (1-x^2) is...

I have 1/ SQRT(1-x^2) for the regualkr derivative but I am not sure how "for absolute value of (x) < 1" would change the answer if it does.
Click to expand...
1) Don't worry too much about the absolute value in this case. It is only defining the Domain. x = 1 and x = -1 are no good. x > 1 and x < -1 are no good. |x| < 1 keeps you away from the bad places.

2) You had better try that derivative again. You seem to have forgotten the Chain Rule - a couple of times.
 
Hello, xc630!

Here's #2 . . .

2) The tangent line to the graph \(\displaystyle y\:=\:e^{2-x}\) at the point (1, e) intersects both coordinate axes.
What is the area of the triangle formed by this tangent line and the coordinate axes?
Click to expand...

We will find the equation of that tangent.

We have a point on the tangent \(\displaystyle (1,\,e)\)
We need the slope of the tangent.

The slope of a tangent is given by the derivative. .\(\displaystyle y'\:=\:-e^{2-x}\)
. . At \(\displaystyle (1,\,e)\), the slope is: \(\displaystyle \:m \:=\:-e\)

The equation of the tangent is: \(\displaystyle \:y\,-\,e\:=\:-e(x\,-\,1)\;\;\Rightarrow\;\;y \:=\:-ex \,+\,2e\)


For the x-intercept, let \(\displaystyle y\,=\,0:\;-ex\,+\,2e\:=\:0\;\;\Rightarrow\;\;x\,=\,\ln(2e)\)
. . x-intercept: \(\displaystyle \,(\ln2e,\,1)\)

For the y-intercept, let \(\displaystyle x\,=\,0:\;y\:=\:-e^0\,+\,2e\;\;\Rightarrow\;\;y\:=\;2e\,-\,1\)
. . y-intercept: \(\displaystyle \,(0,\,2e\,-\,1)\)


Make a quick sketch . . .
The base of the triangle is: \(\displaystyle \,ln(2e)\)
The height of the triangle is: \(\displaystyle \,2e\,-\,1\)

Area of triangle: \(\displaystyle \:A\:=\:\frac{1}{2}\cdot\ln(2e)\cdot(2e\,-\,1)\)

 
You can simplify your life with properties of logarithms.

log(sqrt(f(x))) = (1/2)log(f(x))
 
Hello soroban thanks for the help. I understand your logic but I am not sure how you calcaulated the x and y intercepts. When y= 0 doesn't x = 2? Where did
ln (2e) come from? Not seeing how you got the y-intercept too. Thanks
 
I think I got it soroban. From the equation you derived the y intercept is (0, 2e) and plugging in y=0 the x-intercept is (2,0)

Then the area = 1/2 * (2*2e)= 2e
 
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