The equation of the tangent to the curve y = xsinx at the point (pi/2, pi/2) is..?

Question

SHOW ALL WORK!!

Answer choices are...

A. y = x - pi
B. y = pi/2
C. y = pi - x
D. y = x + Pi/2
E. y = x

Answer 1

y = x sinx

differentiate

y' = xcosx + sinx

The derivative of y is slope of tangent

so slope of tangent, m = xcosx + sinx

at x = pi/2, the slope is

y'(pi/2) = (pi/2)cos(pi/2) + sin(pi/2)

=>0 + 1

so slope of tangent is 1

The equation of tangent with slope 1 and containing point(pi/2,pi/2)

is

y - pi/2 = 1(x-pi/2)

y = x

Answer 2

Use the product rule and differentiate y=xsinx
You will get y'=sinx+xcosx
Plug in x=pi/2 and get y'=1
Since derivative also means the slope of the point. So the slope at x=pi/2 is 1.
(y-pi/2)/(x-pi/2)=1
So y-pi/2=x-pi/2
Simplify and get y=x
E is answer.

Answer 3

find derivative

y' = d/dx(x) (sinx) + d/dx(sinx) (x)
y' = sin(x) + xcos(x)

plug pi/2 in for x
y' = sin(pi/2) + (pi/2) cos(pi/2)
y' = 1 + (pi/2)(0)
y' = 1

so the slope of the tangent line is 1

y = mx + b
pi/2 = 1(pi/2) + b
pi/2 = pi/2 + b
b = 0

y = x <== answer

Answer 4

y = xsinx

Find y' so we can get the slope when x=pi/2
Using the product rule, we get
y' = sinx + xcosx

At x= pi/2
y' = sinx + xcosx
y' = sin(pi/2) + (pi/2)cos(pi/2)
y' = 1 + (pi/2)(0) = 1 = m

Equation for tangent lines is y=mx+b
Using the point (pi/2,pi/2), we get
(pi/2)= (1)(pi/2) +b
b=0

So your equation of the tangent line is y=(1)x or simply y=x
[Answer: E]