the derivative of tangent dy/dx [ tan(u) ]= [sec^(2)u]u' this means that the derivative of tangent of u is secant squared u times the derivative of u.
Yes, the derivative of an equation is the slope of a line tangent to the graph.
Take the derivative of the function.
-ln|cos x| + C
Calculus! So, when you differentiate a function or find the derivative, as it is also called, you are finding the rate of change of the function. An easy way to find the derivative at a certain point on a function is to draw a line tangent to the function at that point; the slope of the tangent line is the derivative.
Why: Because that's what the derivative means, the way it is defined - the slope of the curve at any point of the line.
The derivative at any point in a curve is equal to the slope of the line tangent to the curve at that point. Doing it in terms of the actual expression of the curve, find the derivative of the curve, then plug the x-value of the point into the derivative to find the derivative at that point.
In order to find the equation of a tangent line you must take the derivative of the original equation and then find the points that it passes through.
Because it leads to the limit concept which in turn leads to concept of derivative...
The rate of change on that line. This is called the tangent and is used in the application of the derivative.
Regardless of what 'x' is, (x)0 = 1 . tan(1 radian) = 1.55741 (rounded) tan(1 degree) = 0.01745 (rounded) We can't remember the derivative of the tangent right now, but it doesn't matter. This particular tangent is a constant, so its derivative is zero.
d/dx[ tan-1(x) ] = 1/(1 + x2)
The answer will depend on the context. If the curve in question is a differentiable function then the gradient of the tangent is given by the derivative of the function. The gradient of the tangent at a given point can be evaluated by substituting the coordinate of the point and the equation of the tangent, though that point, is then given by the point-slope equation.