Find an equation of the tangent line/differntiation

[getoutofmylaboratory]

Find an equation of the tangent line to y = f(x) at x = a

f(x) = x sin x, a= pi/2

f(a) = f(pi/2) = pi/2 sin pi/2 = pi/2

So from here, I think I need to find f'(pi/2), but not sure how to do that. And once found, how would I use that and the answer of (pi/2) above to find the equation?

Thanks

 

[getoutofmylaboratory]

I think you'll find the most useful formula for a line here to be y(x) = f(a) + f'(a)(x-a)
This intercepts f at the point a and has slope equal to the derivative of f at a.

I'm going to guess that you should be able to find the derivative of x sin(x).

Give this a shot and if you really have no idea post back.

remember the product rule for derivatives d/dx(f(x) g(x)) = f'(x) g(x) + f(x) g'(x)

Thanks for the reply. I know that the derivative of x sin(x) = sin(x) + x cos(x). So, does that mean this becomes sin(pi/2) + (pi/2) cos(pi/2)?

 

[getoutofmylaboratory]

yes and yes

Ok, so this gives me sin(pi/2) + (pi/2) cos(pi/2) = 1

So, from here: y(x) = f(a) + f'(a)(x - a) = (pi/2) + 1 * (x - pi/2)? What is x here?

 

[getoutofmylaboratory]

x is your independent variable that let's you move on the line. maybe a drawing is called for

View attachment 3497

x is just the independent variable for both f[x] and the line l[x] = f[a] + f'[a](x-a)

Ok, so then the equation of the tangent line is just y(x) = (pi/2) + 1 * (x - pi/2). Or am I totally missing the ball here?

 

[getoutofmylaboratory]

you've got it.

Awesome, thanks for your help.