# For the slope of the line at a point, why am I getting a different result by using the calculus method?

I am evaluating the slope of the secant as it approaches $f(30)$ for the function $f(x) = 2\sin(x) - 2$.

Using calculus I can easily find that the derivative is $f'(x) = 2\cos(x)$.

If I sub in $30$ degrees for $x$ I find that the slope of the tangent at $f(30)$ is approximately $1.73205$.

However, using the slope of the secant method I calculate

$$\operatorname{slope} = \frac{f(30.001) - f(30)}{0.001}\$$

Which gives me $0.03023$. I'm quite confused here as to why. I did double check that all of the calculations are in degrees. I can't understand the problem here.

• Calculus works with radians, not degrees. – Gerry Myerson Jun 8 '13 at 1:27

Try converting $30^\circ = \pi/6\;\text{radians}$, and evaluate $f'(x)$ at $x = \pi/6$
The crucial point is that $\lim_{x\to 0} \frac{sin(x)}{x}= 1$, important in finding the derivative of sin(x), is true only when x is in radians. With x in degrees, that limit would be $\frac{\pi}{180}$ which is approximately 0.01745.