d/dx csc^-1(u)
To find the derivative of csc^-1(u) with respect to x, we can use the chain rule
To find the derivative of csc^-1(u) with respect to x, we can use the chain rule. Let’s start by first expressing csc^-1(u) in terms of an inverse trigonometric function.
The inverse cosecant function, csc^-1(u), is defined as the angle whose cosecant is equal to u. In other words, if y = csc^-1(u), then u = csc(y).
Now, let’s find the derivative of csc(y) with respect to y, and then multiply it by dy/dx to find the derivative of csc^-1(u) with respect to x.
1. Differentiating csc(y) with respect to y:
To find the derivative of csc(y), we can rewrite it as 1/sin(y). Applying the quotient rule, we have:
d/dy [csc(y)] = d/dy [1/sin(y)] = (0 * sin(y) – 1 * cos(y))/sin^2(y) = -cot(y)/sin^2(y)
2. Multiplying by dy/dx:
Now, we need to multiply d/dy [csc(y)] by dy/dx to find the derivative of csc^-1(u) with respect to x.
We know that u = csc(y), so we can rewrite it as y = csc^-1(u). Taking the derivative of both sides with respect to x, we get:
(d/dx)y = (d/dx)csc^-1(u)
dy/dx = (d/du)csc^-1(u) * (du/dx)
Since dy/dx = d/dx y = d/dx csc^-1(u), we can substitute this back into the equation:
d/dx csc^-1(u) = (d/du)csc^-1(u) * (du/dx)
3. Finding (d/du)csc^-1(u):
To find (d/du)csc^-1(u), we can differentiate the expression y = csc^-1(u) implicitly with respect to u:
dy/du = (d/du) csc^-1(u)
Now, we need to find dy/du. Since y = csc^-1(u) and u = csc(y), we have:
u = csc(y) = 1/sin(y)
To differentiate both sides with respect to u, we use the chain rule. Let’s denote sin(y) as sin(g(u)):
du/dy * dy/dx = -1/sin^2(y) * dsin(g(u))/du
Where g(u) is an intermediate function. By rearranging and isolating dy/dx, we get:
dy/dx = -1/(du/dy * sin^2(y)) * dsin(g(u))/du
Notice that du/dy is the reciprocal of dy/du, so we can rewrite it as:
dy/dx = -1/(dy/du * sin^2(y)) * dsin(g(u))/du
Remember that we are looking for (d/du) csc^-1(u), which can be expressed in terms of sin(g(u)), so let’s solve for this term:
sin(g(u)) = 1/u
Differentiating both sides with respect to u using the chain rule, we get:
cos(g(u)) * dg(u)/du = -1/u^2
Solving for dg(u)/du, we have:
dg(u)/du = -1/(u * cos(g(u)))
Now, we have all the pieces to compute d/dx csc^-1(u):
d/dx csc^-1(u) = (d/du)csc^-1(u) * (du/dx)
= -1/(dy/du * sin^2(y)) * dsin(g(u))/du
= -1/(dy/du * sin^2(y)) * (-1/(u * cos(g(u))))
Finally, substituting in the expressions for dy/du and dsin(g(u))/du:
d/dx csc^-1(u) = -1/(dy/du * sin^2(y)) * (-1/(u * cos(g(u))))
= -1/((-1/(dy/du * sin^2(y))) * (u * cos(g(u))))
= 1/(dy/du * sin^2(y) * u * cos(g(u)))
Therefore, the derivative of csc^-1(u) with respect to x is given by 1/(dy/du * sin^2(y) * u * cos(g(u))).
Please note that this expression may seem complex, but it is the result of using the chain rule and the implicit differentiation technique to find the derivative of an inverse trigonometric function.
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