Q: What is the derivative of tan?

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If it's tan(x3), the the thingy is 3x2/(cos2(x3)) i think Else, if it's tan(3x), then it's 3/(cos2x) If you need more help with this, search "chain rule" and "derivative" on wikipedia.

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 sec(2x) = 2sec(2x)tan(2x)

Given y = tan x: dy/dx = sec^2 x(secant of x squared)

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The derivative of tan(x) is sec2(x).(Which is the same as 1/cos2(x).

The derivative of sec(x) is sec(x) tan(x).

If it's tan(x3), the the thingy is 3x2/(cos2(x3)) i think Else, if it's tan(3x), then it's 3/(cos2x) If you need more help with this, search "chain rule" and "derivative" on wikipedia.

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.

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.

sec(x)tan(x)

d/dx csc(x) = - csc(x) tan(x)

In this case, you'll need to apply the chain rule, first taking the derivative of the tan function, and multiplying by the derivative of 3x: y = tan(3x) ∴ dy/dx = 3sec2(3x)

d/dx sec(2x) = 2sec(2x)tan(2x)

d/dx(arctan x) = X = 1/(1 + x2)

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

The derivative of the natural log is 1/x, therefore the derivative is 1/cos(x). However, since the value of cos(x) is submitted within the natural log we must use the chain rule. Then, we multiply 1/cos(x) by the derivative of cos(x). We get the answer: -sin(x)/cos(x) which can be simplified into -tan(x).