anonymous
  • anonymous
Find the equation of the tangent line to the graph of y=x^sinx at the point (π/2, π/2).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
First we need to find y' (or d/dx) for this equation. Do you know the formula for finding the derivative of this function?
SolomonZelman
  • SolomonZelman
Use logarithmic differentiation to find this derivative.
SolomonZelman
  • SolomonZelman
Should I give an example of doing this?

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anonymous
  • anonymous
No I do not
WYKEISE
  • WYKEISE
want us to give you the answer
anonymous
  • anonymous
\[\huge x^{\sin(x)}=e^{\sin(x)\ln(x)}\] differentiate that using chain and product rule
anonymous
  • anonymous
@mgeorge20 Until now to find the derivative of \[x^{n}\] we used \[nx^{n-1}\] where n = number However, this is not raised to n (meaning a number) it is raised to sin(x) so we first need to take the natural log of both sides
anonymous
  • anonymous
So first we re-write as ln y = ln x^sin x Does this make sense so far?
anonymous
  • anonymous
We take the natural log because it is raised to sinx?
anonymous
  • anonymous
not to butt in, but the definition of \[x^{\sin(x)}\]is \[e^{\sin(x)\ln(x)}\]
anonymous
  • anonymous
Correct if \[y =x^{x}\] We take the ln of both sides and then we use the rule for d/dx of ln(u) = 1/u * u'
lochana
  • lochana
yes. use logarithms \[y=x^{sinx}\]\[lny = sinx.lnx\]
anonymous
  • anonymous
@LeibyStrauss okay makes sense so far
anonymous
  • anonymous
Next using the log power rule rewrite as ln y = sinx ln x Next take derivative of both sides. Let me know if you want me to do it or if you're ok with doing it.
lochana
  • lochana
@mgeorge20 can you do it now?
anonymous
  • anonymous
@LeibyStrauss so the six comes down and then you multiply it with the natural log, ok.
anonymous
  • anonymous
Now you get y by itself? @LeibyStrauss
anonymous
  • anonymous
I'm not sure where the six came from. d/dx(ln y = sinx ln x) => d/dx ln y = d/dx sinx ln x => d/dx ln y = ln 1/y y' For d/dx sinx ln x we need to use the product rule d/dx sinx ln x = sin(x) * d/dx(lnx) + ln(x) * d/dx(sinx)
anonymous
  • anonymous
I think you mean sin(x)
anonymous
  • anonymous
@LeibyStrauss yeah, sorry. So since the sinx came down, you multiply that with natural log x Then you do the product rule which means to take the derivative of each?
anonymous
  • anonymous
Correct
anonymous
  • anonymous
1/y y' = [sinx * 1/x + cos(x) ln(x)] = 1/y y' = [(sinx/x) + cos(x) ln(x)] Now we want to isolate y' so we multiply both sides by y
anonymous
  • anonymous
y' = y [(sinx/x) + (cos(x) ln(x))] y' = y[sinx/x] + y[cos(x) ln(x)] Since we substitute the x value and y value in
anonymous
  • anonymous
Let me know if I should clarify anything or when you're ready for the next step
anonymous
  • anonymous
Okay I'm ready for the next step
anonymous
  • anonymous
y' = y[sinx/x] + y[cos(x) ln(x)] x = pi/2 y = pi/2 At this point we plug in the x and y value y' = pi/2[sin(pi/2)/(pi/2)] + y[cos(pi/2) ln(pi/2)] sin(pi/2) = 1 cos(pi/2) = 0 \[y' = \frac{ \pi }{ 2 }(\frac{ 1 }{ \frac{ \pi }{ 2 } })+ 0[\ln(\pi/2)]\] \[y' = \frac{ \pi }{ 2 }*\frac{ 2 }{ \pi }\] So y' = 1 y' = slope = m
anonymous
  • anonymous
@LeibyStrauss okay
anonymous
  • anonymous
We have one final step, we need to write the equation in the form of of y = mx + b. we have "m" (m = 1) we need to find b So we plug in the y and x value again to solve for b pi/2 = 1(pi/2) + b pi/2 = pi/2 + b subtract pi/2 from both sides b = 0 y = 1x + 0 the final answer is y = x
anonymous
  • anonymous
Let me know if you have any questions or if you want to work on the next problem
anonymous
  • anonymous
Next problem

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