Question

Find the equation of the tangent line to the graph of y=x^sinx at the point (π/2, π/2).

Answer 1

First we need to find y' (or d/dx) for this equation. Do you know the formula for finding the derivative of this function?

Answer 2

Use logarithmic differentiation to find this derivative.

Answer 3

Should I give an example of doing this?

Answer 4

No I do not

Answer 5

want us to give you the answer

Answer 6

\[\huge x^{\sin(x)}=e^{\sin(x)\ln(x)}\] differentiate that using chain and product rule

Answer 7

@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

Answer 8

So first we re-write as ln y = ln x^sin x

Does this make sense so far?

Answer 9

We take the natural log because it is raised to sinx?

Answer 10

not to butt in, but the definition of \[x^{\sin(x)}\]is \[e^{\sin(x)\ln(x)}\]

Answer 11

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'

Answer 12

yes. use logarithms \[y=x^{sinx}\]\[lny = sinx.lnx\]

Answer 13

@LeibyStrauss okay makes sense so far

Answer 14

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.

Answer 15

@mgeorge20 can you do it now?

Answer 16

@LeibyStrauss so the six comes down and then you multiply it with the natural log, ok.

Answer 17

Now you get y by itself? @LeibyStrauss

Answer 18

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)

Answer 19

I think you mean sin(x)

Answer 20

@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?

Answer 21

Correct

Answer 22

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

Answer 23

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

Answer 24

Let me know if I should clarify anything or when you're ready for the next step

Answer 25

Okay I'm ready for the next step

Answer 26

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

Answer 27

@LeibyStrauss okay

Answer 28

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

Answer 29

Let me know if you have any questions or if you want to work on the next problem

Answer 30

Next problem