- #1

- 2

- 0

## The Attempt at a Solution

lny = cosx(ln(ln(x))) d/dx

= -sinx(ln(ln(x))) + cosx/(ln(x)(x))

y' = y(-sinx(ln(ln(x))) + cosx/(ln(x)(x)))

this is the part where I get stuck

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter dvaughn
- Start date

- #1

- 2

- 0

lny = cosx(ln(ln(x))) d/dx

= -sinx(ln(ln(x))) + cosx/(ln(x)(x))

y' = y(-sinx(ln(ln(x))) + cosx/(ln(x)(x)))

this is the part where I get stuck

- #2

jhae2.718

Gold Member

- 1,170

- 20

## The Attempt at a Solution

lny = cosx(ln(ln(x))) d/dx

= -sinx(ln(ln(x))) + cosx/(ln(x)(x))

y' = y(-sinx(ln(ln(x))) + cosx/(ln(x)(x)))

this is the part where I get stuck

When you get to

[itex]y'=y\left[ {\frac{\cos\left(x\right)}{x\ln\left(x\right)}-\sin\left(x\right)\ln\left(\ln\left(x\right)\right)}\right][/itex]

substitute [itex]\ln\left(x\right)^{\cos\left(x\right)}[/itex] for y.

Then take [itex]y'\left(\frac{\pi}{2}\right)[/itex] as the slope of your tangent line and find what y-intercept will put the line through [itex]\left(\frac{\pi}{2},1\right)[/itex].

Last edited:

Share: