d/dx sec^-1(u)
To find the derivative of the inverse secant function, we can first start by writing it in terms of the natural logarithm function
To find the derivative of the inverse secant function, we can first start by writing it in terms of the natural logarithm function.
The inverse secant function, denoted as sec^-1(u) or arcsec(u), is defined as the angle whose secant is equal to u. In other words:
arcsec(u) = θ, where sec(θ) = u.
To find the derivative of sec^-1(u) with respect to x (d/dx sec^-1(u)), 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, we have y = sec^-1(u), and we want to find dy/dx.
Let’s start by letting θ = sec^-1(u), so sec(θ) = u.
Now, we can find the derivative of both sides of the equation sec(θ) = u with respect to x:
d/dx [ sec(θ) ] = d/dx [ u ].
Using the Chain Rule, the left side becomes:
sec(θ) * tan(θ) * dθ/dx.
Since θ = sec^-1(u), we can rewrite this as:
u * tan(sec^-1(u)) * dθ/dx.
Now, to find d/dx sec^-1(u), we need to solve for dθ/dx.
From the definition of the inverse secant, sec(θ) = u, we can apply the reciprocal property of secant and obtain:
cos(θ) = 1/u.
Taking the derivative of both sides with respect to x:
d/dx [ cos(θ) ] = d/dx [ 1/u ].
Using the Chain Rule again, we get:
-sin(θ) * dθ/dx = -1/u^2 * du/dx.
We can rewrite this as:
dθ/dx = -1/(u * sin(θ)) * du/dx.
Substituting this expression back into our previous equation:
d/dx sec^-1(u) = u * tan(sec^-1(u)) * (-1/(u * sin(θ)) * du/dx).
Simplifying this further, we have:
d/dx sec^-1(u) = -tan(sec^-1(u))/(u * sin(θ)) * du/dx.
From the trigonometric identity sec(θ) = u, we have:
θ = cos^-1(1/u).
Therefore, substituting back:
d/dx sec^-1(u) = -tan(cos^-1(1/u))/(u * sin(cos^-1(1/u))) * du/dx.
This expression gives the derivative d/dx sec^-1(u) in terms of u and its derivative du/dx.
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