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Question
Working with special derivatives, don't understand this practice problem.

Find an equation of the line tangent to the graph of:
y=sin(x)+5 at x=pi

Thank you for your time and assistance.

Answer
The graph of y = sin(x) +5 is pretty simple and I'm guessing you can either graph it by hand or have a calculator that can do it. In any case, the graph looks like a regular sine wave shifted up by 5 units. So, for instance, at x = 0, sin(x) = 0 so y(0) = 5.

The question asks for the equation of the line tangent to this curve at a particular point. The equation for a line consists of a slope and intercept, which can be written in the familiar way

z = mx +b   m = slope and b = intercept.

I'm using the letter z for the line so as not to confuse it with the expression for y. Now, the slope of the tangent at a point along a curve is just the value of the first derivative of the function defining the curve, evaluated at the point. So we need

dy/dx = d(sin(x) + 5)/dx = cos(x) at x = pi, or

dy/dx = cos(pi) = -1.

This is our slope. The intercept is defined as the value of y at x = 0, so in the present case, y(0) = sin(0) + 5 = 5.

The equation of the tangent line is thus

z = -x + 5.

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