## -csc(x)cot(x) is the derivative of?

### To find the derivative of the function -csc(x)cot(x), we can use the product rule and the chain rule for differentiation

To find the derivative of the function -csc(x)cot(x), we can use the product rule and the chain rule for differentiation.

Let’s start by rewriting the function using trigonometric identities:

-csc(x)cot(x) = -(1/sin(x)) * (cos(x)/sin(x))

Now, using the product rule, whose formula is (uv)’ = u’v + uv’, we have:

(uv)’ = (-1/sin(x))’ * (cos(x)/sin(x)) + (-1/sin(x)) * (cos(x)/sin(x))’

To find the derivative of (-1/sin(x)), we can use the quotient rule, which states that (u/v)’ = (u’v – uv’)/v^2. In this case, u = -1 and v = sin(x). Differentiating, we get:

(-1/sin(x))’ = (0 * sin(x) – (-1) * cos(x))/sin^2(x) = cos(x)/sin^2(x)

Now, for the derivative of (cos(x)/sin(x)), we use the quotient rule again. Let u = cos(x) and v = sin(x). Differentiating, we have:

(cos(x)/sin(x))’ = (sin(x)(-sin(x)) – cos(x)cos(x))/(sin(x))^2 = (-sin^2(x) – cos^2(x))/sin^2(x)

But we know that sin^2(x) + cos^2(x) = 1, so we can simplify further:

(-sin^2(x) – cos^2(x))/sin^2(x) = (-1)/sin^2(x) = -csc^2(x)

Substituting these derivatives back into our earlier expression, we get:

(-csc(x)cot(x))’ = (cos(x)/sin^2(x)) * (cos(x)/sin(x)) + (-1/sin(x)) * (-csc^2(x))

Simplifying further, we have:

= (cos^2(x))/(sin^3(x)) + csc^2(x)/sin(x)

= cot^2(x)/sin(x) + csc^2(x)/sin(x)

Now, combining the terms by finding a common denominator, we get:

= (cot^2(x) + csc^2(x))/sin(x)

Therefore, the derivative of -csc(x)cot(x) is (cot^2(x) + csc^2(x))/sin(x).

##### More Answers:

The Chain Rule | Finding the Derivative of sec²(x) using the Quotient and Chain RuleUnderstanding the Derivative of csc²(x) and its Application in Calculus

Derivative of tan(x)sec(x) | The Product Rule in Differentiation Explained