Question

Find an equation of he tangent line to the curve at the given point

y= sinx+sin^2 x , (0,0)

Answer 1

y' = cos(x) + cos(x)sin(x) + sin(x)cos(x)
y' = cos(x) + 2cos(x)sin(x)

At (0,0), y' = cos(0) + 2cos(0)sin(0) = 1

There's your slope. Now for the equation,

y + 0 = 1(x+0)
y=x

Answer 2

Well, to find the slope of an equation at any point in the function, we first have to find the derivative of the function. The derivative of y = sin(x) + sin^2(x) is y' = cos(x) + 2cos(x)sin(x). To find the slope at a specific point, you just plug in the x-value of the point into the derivative. I plugged in zero, and got the slope, then used point-slope formula to get the equation of the line.

What exactly don't you understand?

Answer 3

its finding the derivatives taht i find diffivcult to do

Answer 4

Okay. We have

y = sin(x) + sin^2(x)

In order to find the derivative of this function, we must know that d/dx [sin(x)] = cos (x).

Then, we must know that d/dx [f(x) + g(x)] = d/dx [f(x)] + d/dx [g(x)].

And also the product rule: d/dx [f(x) * g(x)] = f'(x)g(x) + f(x)g'(x)

So, y' = d/dx [sin(x) + sin^2(x)]

y' = d/dx [sin(x)] + d/dx [sin^2(x)]
The derivative of sin(x) is cos(x).
y' = cos(x) + d/dx [sin^2(x)]
Now, let's deal with sin^2(x). This can be expanded to sin(x) * sin(x).

Product rule. d/dx [ sin(x)*sin(x) ] = d/dx [sin(x)]sin(x) + d/dx [sin(x)]sin(x) = cos(x)sin(x) + cos(x)sin(x) = 2cos(x)sin(x).

Answer 5

thankls foir breaking ut down....it makes it easier b.c wen i try to undertans math n steps are skipped it confuses me...srrry im just slow in math