# Thread: Equation of the tangent line

1. ## Equation of the tangent line

Find an equation of the tangent line to the curve at the given point. Leave your answer solved for y.

Y=(1+x)cosx (0,1)

Okay, so I think you just take the derivative of the function, but how do I do that and make the rest of the problem work?

2. The derivative is $y'=\cos(x)-(1+x)\sin(x)$.

3. Originally Posted by bobsanchez
Find an equation of the tangent line to the curve at the given point. Leave your answer solved for y.

Y=(1+x)cosx (0,1)

Okay, so I think you just take the derivative of the function, but how do I do that and make the rest of the problem work?
Why did you suspect to take the derivative ?

The derivative gives the slope of the tangent.
The tangent is a straight line.
The point (0,1) is x=0, y=1 corresponding to a point on the tangent, the point of tangency to the curve.

To make the rest of the problem work......
How do you write the equation of a straight line?
If you have 2 fixed points, you can do it.
If you have a fixed point and the slope you can do it.

What do you do here?

4. Is the answer y = x + 1?

5. Originally Posted by bobsanchez
Is the answer y = x + 1?
Yes.

The derivative at (0,1) is 1, so the tangent slope is 1.

The equation of the tangent is y-1=m(x-0), so y-1=x... y=x+1