What is the derivative of tangent?

Updated: 4/28/2022
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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.

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Q: What is the derivative of tangent?
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Derivative as the Slope of a Tangent?

Yes, the derivative of an equation is the slope of a line tangent to the graph.

How do you find the slope of a tangent?

Take the derivative of the function.

What is the anti-derivative of a tangent function?

-ln|cos x| + C

Why does taking the derivative of a function give you the slope of the tangent?

Why: Because that's what the derivative means, the way it is defined - the slope of the curve at any point of the line.

What is the derivative of tan x0 its x to the power 0?

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.

What does the derivative at a point mean?

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.

How do you find the equation of a tangent line?

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.

What does the slope graph tells?

The rate of change on that line. This is called the tangent and is used in the application of the derivative.

Why are tangent lines important?

Because it leads to the limit concept which in turn leads to concept of derivative...

What is the derivative of inverse tangent of x?

d/dx[ tan-1(x) ] = 1/(1 + x2)

How do you get the tangent line without the graph?

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.

What is the square-root of a tangent?

The tangent is essentially the derivative of the function. The square-root is just what ever function that is takes two of that function to equal the tangent. If you need further help on this question just send me a message on my message board and id be glad to help you out.