d/dx [tan x]
To find the derivative of tan x, denoted as d/dx[tan x], we can use the quotient rule
To find the derivative of tan x, denoted as d/dx[tan x], we can use the quotient rule.
The quotient rule states that for two functions u(x) and v(x),
(d/dx)[u(x)/v(x)] = (v(x)*[du(x)/dx] – u(x)*[dv(x)/dx]) / [v(x)]^2
In this case, u(x) = 1 and v(x) = cot x.
We can rewrite tan x as 1/cot x, since tan x is the same as 1 divided by cot x.
Using the quotient rule, we have:
(d/dx)[tan x] = (cot x * [d(1)/dx] – 1 * [dcot x/dx]) / [cot x]^2
Now, let’s evaluate the individual derivatives:
[d(1)/dx] is the derivative of the constant function 1, which is 0.
[dcot x/dx] is the derivative of cot x. To find this, we can use the chain rule. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).
In this case, f(x) = cot x and g(x) = x. Taking the derivative of cot x with respect to x gives us -csc^2 x. Since g(x) = x, g'(x) = 1.
Therefore, [dcot x/dx] = -csc^2 x * 1 = -csc^2 x.
Now, plugging these values back into the quotient rule equation:
(d/dx)[tan x] = (cot x * 0 – 1 * (-csc^2 x)) / [cot x]^2
= csc^2 x / [cot x]^2
Simplifying further, we can rewrite csc^2 x as 1/sin^2 x and cot x as cos x / sin x:
= (1/sin^2 x) / [(cos x / sin x)^2]
= (1/sin^2 x) / (cos^2 x / sin^2 x)
= 1 / cos^2 x
= sec^2 x
Therefore, the derivative of tan x, d/dx[tan x], is equal to sec^2 x.
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