Find an equation of the tangent line to the curve at the given point.
y = sin(sin x), (pi, 0)
Nothing simpler...
y = sin(sin x)
y' = cos(sin x)*cos x
y'(pi) = cos(0)*cos(pi) = 1(-1) = -1
so, you want the line through (pi,0) with slope = -1
(y-0)/(x-pi) = -1
y = pi - x
To find the equation of the tangent line to the given curve at the point (π, 0), we need to determine the slope of the tangent line at that point.
The slope of the tangent line at a given point on a curve can be found by taking the derivative of the function and evaluating it at that point.
Let's find the derivative of y = sin(sin x) using the chain rule.
First, let's take the derivative of the outer function sin(sin x) with respect to the inner function sin x.
Derivative of sin(sin x) with respect to sin x = cos(sin x)
Next, we need to multiply this by the derivative of the inner function sin x with respect to x.
Derivative of sin x with respect to x = cos x
Now, let's multiply these two derivatives together to get the derivative of y = sin(sin x) with respect to x:
dy/dx = cos(sin x) * cos x
To find the slope of the tangent line at the point (π, 0), we need to evaluate dy/dx at x = π:
dy/dx = cos(sin π) * cos π
Since sin π = 0 and cos π = -1, we have:
dy/dx = cos(0) * (-1) = -1
Now that we have the slope of the tangent line, we can use the point-slope form of a line to find its equation.
The point-slope form of a line is given by:
y - y₁ = m(x - x₁)
where (x₁, y₁) is a point on the line and m is the slope of the line.
Plugging in the values x₁ = π, y₁ = 0, and m = -1, we get:
y - 0 = -1(x - π)
Simplifying this equation, we have:
y = -x + π
Therefore, the equation of the tangent line to the curve y = sin(sin x) at the point (π, 0) is y = -x + π.