Find equation of tangent line

  • Thread starter dvaughn
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Find the equation of the tangent line to the curve y=(lnx)^cosx at the point (pi/2, 1)?

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
 

Answers and Replies

  • #2
jhae2.718
Gold Member
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Find the equation of the tangent line to the curve y=(lnx)^cosx at the point (pi/2, 1)?

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:

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