Find dy/dx

[Tueseve728]

Find dy/dx for the following:

y=ln(2^tanx)


x sin ^-1 y=1+x^2

y=x^x


Are there any points (ordered pairs) on the curve y=x-e^-x where the slope is 2? If so find them.

 

[stapel]

Just to clarify, this is my understanding of your posting:

. . . . .1) y = ln(2^tan(x))

. . . . .2) x sin^-1(y) = 1 + x²

. . . . .3) y = x^x

. . . . .4) Find all points, if any exist, on the curve y = x - e^-x
. . . . .where the slope of the tangent line is m = 2.

Please confirm or correct:

(I know that formatting some of these more-involved exercises can get a bit hairy, so I'd just like to confirm that I'm reading the same thing you're writing. Thank you for your understanding.)

Eliz.

 

[tkhunny]

Find dy/dx for the following:

y=ln(2^tanx)

Wow. This one looks fun. SUPER Chain Rule Zone!

f(x) = ln(x)
g(x) = 2^x
h(x) = tan(x)

f'(x) = 1/x
g'(x) = [2^x]*ln(2)
h'(x) = [sec(x)]^2

Put it all together

y' = (1/(2^tanx))*([2^tan(x)]*ln(2))*[sec(x)]^2

You may wish to simplify that a bit, if possible.

More importantly, I think, after you see what went on, is that you should be proud that you waded through it. Go ahead. Pat yourself on the back.

 

[Tueseve728]

yea that is correct

 

[galactus]

Find dy/dx for the following:

y=ln(2^tanx)


x sin ^-1 y=1+x^2

y=x^x

Use logarithmic differentiation for this one. Take ln of both sides:

ln(y)=xln(x).

Differentiate both sides:

y'/y=ln(x)+1

One more little step. Remember what y equals.

Can you finish up from here?.



Are there any points (ordered pairs) on the curve y=x-e^-x where the slope is 2? If so find them.

 

[stapel]

yea that is correct

Thank you.

Exercises (1) and (3) have already been worked on. For (2), there are probably many ways to go, but I might start out like this:

. . . . .x sin^-1(y) = 1 + x²

. . . . .sin^-1(y) = (1 + x²) / x

. . . . .y = sin[(1 + x²) / x]

All I've done so far is isolate "y". This is just a personal preference, and is not, I don't think, strictly "required". Then:

. . . . .dy/dx = cos[(1 + x²) / x] [(2x² - 1 - x²) / x</sup>2</sup>]

Then simplify.

For (4), differentiate, and set the derivative equal to "2".

. . . . .y = x - e^-x

. . . . .dy/dx = 1 + e^-x

. . . . .1 + e^-x = 2

. . . . .e^-x = 1

What then must x equal? At this x-value, what is the curve's y-value?

Hope this helps a bit. If you get stuck finishing any of the exercises, please reply showing how far you have gotten. Thank you.

Eliz.

 

[Tueseve728]

So for #4 is x=0? so does that mean there are no ordered pairs?


Also for #3, i dont know what to do after that step that was posted.

 

[galactus]

So for #4 is x=0? so does that mean there are no ordered pairs?


Also for #3, i dont know what to do after that step that was posted.

Solve for y'.

 

[Tueseve728]

so is the answer to #3 (1/x)

 

[stapel]

So for #4 is x=0?

That's what I get.

so does that mean there are no ordered pairs?

Why would getting an x-value and plugging into y to get a y-value, so you have (x, y), mean that there is no ordered pair (x, y)? I'm sorry, but I don't understand what you're asking here.

Also for #3, i dont know what to do after that step that was posted.

You were given that y'/y = ln(x) + 1, where y = x^x How did you solve for "y'=" and get "y' = 1/x"? Please reply showing all your steps.

Thank you.

Eliz.

 

[Tueseve728]

This is what I got for number 3.. I'm really confused with derivatives,
ln y=x ln x

=ln(x)+1

d/dx(ln(x)+1)

d/dx(1)+d/dx(ln(x)+1)=d/dx(ln(x))=1/x

Along with #1,
y1=(1/2^tan x)) (ln 2) (sec x^2), that is what i got.....

 

[Tueseve728]

so the answer to #3 is y=(ln x+1)y, then y1=(ln x+1)^x

 

[galactus]

so the answer to #3 is y=(ln x+1)y, then y1=(ln x+1)^x

y'=(ln(x)+1)y, remember what y equals?. y=x^x.

So you have y'=(ln(x)+1)x^x.