Question

How do you find the equation of the tangent line to the curve `y = x sin x` at the point where x = π/2?

Answer 1

`y = x`

Explanation:

Differentiate the function with the product rule.

Differentiation will give you the gradient for the tangent at any point, and you use the product rule whenever a function can be thought of as two functions multiplied together.

If `f(x) = g(x) xx h(x)`
then `f'(x) = g'(x)h(x) + g(x)h'(x)`

so if `y = x xx sinx`
then `dy/dx = 1 xx sinx + x xx cosx = sinx + xcosx`

We know that `x = pi/2`, so the gradient is

`m = sin(pi/2) + pi/2cos(pi/2) = 1 + pi/2 xx 0 = 1`

Therefore, we can say that

`y = mx + c`
`y = (1)x + c`
`y = x + c`

So all we really need to find now is the value for `c`, the `y` intercept. We do this by working out a point `(x,y)` on the graph. We are already given that `x=pi/2`, so

`y = xsinx = pi/2sin(pi/2) = pi/2 xx 1 = pi/2`

`:.`

`(x,y) = (pi/2, pi/2)`

Now we substitute this into the equation we already have for the tangent,

`y = x + c, (x,y)=(pi/2, pi/2)`

`pi/2 = pi/2 + c`

`c = pi/2 - pi/2 = 0`

`:.`

`y = x + c = x + (0) = x`

which means the tangent to the curve `y = xsinx` at `(pi/2, pi/2)` is simply `y = x`.