d/dx cscx
To find the derivative of csc(x) with respect to x, we can use the quotient rule
To find the derivative of csc(x) with respect to x, we can use the quotient rule.
Recall that csc(x) is equal to 1/sin(x). Therefore, we have:
csc(x) = 1/sin(x)
Now, using the quotient rule, the derivative of csc(x) with respect to x will be:
d/dx [1/sin(x)] = (sin(x)d/dx [1] – 1d/dx [sin(x)]) / (sin(x))^2
Since the derivative of a constant (1) is 0, the first part of the numerator simplifies to 0. The derivative of sin(x) with respect to x is cos(x), so the second part of the numerator simplifies to -cos(x). Thus, we have:
d/dx [1/sin(x)] = -cos(x) / (sin(x))^2
Therefore, the derivative of csc(x) with respect to x is -cos(x) / (sin(x))^2.
Alternatively, you can rewrite csc(x) as sin^(-1)(x) and use the chain rule to find its derivative:
Let y = csc(x) = sin^(-1)(x)
Taking the natural logarithm (ln) of both sides:
ln(y) = ln(sin^(-1)(x))
Differentiating both sides with respect to x:
(d/dx) [ln(y)] = (d/dx) [ln(sin^(-1)(x))]
Using the chain rule on the left side:
(1/y) * (d/dx) [y] = (d/dx) [ln(sin^(-1)(x))]
Since y = csc(x), we know that:
(d/dx) [y] = (d/dx) [csc(x)]
So we have:
(1/y) * (d/dx) [csc(x)] = (d/dx) [ln(sin^(-1)(x))]
Plugging in y = csc(x):
(1/csc(x)) * (d/dx) [csc(x)] = (d/dx) [ln(sin^(-1)(x))]
Recall that ln(a^b) = b * ln(a), and in this case, a = sin(x) and b = -1:
(1/csc(x)) * (d/dx) [csc(x)] = (d/dx) [-ln(sin(x))]
Using the derivative of ln(u) with respect to u, where u = sin(x):
(1/csc(x)) * (d/dx) [csc(x)] = -[(d/dx) [sin(x)] / sin(x)]
The derivative of sin(x) with respect to x is cos(x), so we have:
(1/csc(x)) * (d/dx) [csc(x)] = -[cos(x) / sin(x)]
Since csc(x) is equal to 1/sin(x), we can multiply the right side by sin(x)/sin(x) to simplify:
(1/csc(x)) * (d/dx) [csc(x)] = -[cos(x) / sin(x)] * (sin(x) / sin(x))
Simplifying further:
(1/csc(x)) * (d/dx) [csc(x)] = -cos(x) / (sin(x))^2
Therefore, the derivative of csc(x) with respect to x is -cos(x) / (sin(x))^2, which matches the result we obtained using the quotient rule.
More Answers:
Unveiling the Derivative of cot(x) using the Quotient Rule and Trigonometric IdentitiesHow to Find the Derivative of Tan(x) using the Quotient Rule and Simplifying
How to Find the Derivative of sec(x) Using the Quotient Rule