Find dy/dx y=(cot(x)-csc(x))^-1
The question is asking you to calculate the derivative of the function y with respect to x, where y is the inverse of the function cot(x) minus csc(x). In mathematical terms, you're being asked to apply differentiation rules to the function y = (cot(x) - csc(x))^(-1) to find the rate at which y changes as x changes. This involves using the chain rule, quotient rule, and the derivatives of cotangent and cosecant functions to compute dy/dx.
$y = \left(\left(\right. cot \left(\right. x \left.\right) - csc \left(\right. x \left.\right) \left.\right)\right)^{- 1}$
Answer
Solution:
Step 1:
Take the derivative of both sides with respect to $x$: $\frac{d}{dx}(y) = \frac{d}{dx}\left((\cot(x) - \csc(x))^{-1}\right)$.
Step 2:
The derivative of $y$ with respect to $x$ is denoted as $\frac{dy}{dx}$.
Step 3:
Compute the derivative of the right-hand side.
Step 3.1:
Apply the chain rule, which is expressed as $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$, where $f(x) = x^{-1}$ and $g(x) = \cot(x) - \csc(x)$.
Step 3.1.1:
Let $u = \cot(x) - \csc(x)$. Then, differentiate $u^{-1}$ with respect to $u$ and $\cot(x) - \csc(x)$ with respect to $x$: $\frac{d}{du}(u^{-1}) \cdot \frac{d}{dx}(\cot(x) - \csc(x))$.
Step 3.1.2:
Use the power rule, which states that $\frac{d}{du}(u^n) = nu^{n-1}$ for $n = -1$: $-u^{-2} \cdot \frac{d}{dx}(\cot(x) - \csc(x))$.
Step 3.1.3:
Substitute $u$ back into the expression: $-(\cot(x) - \csc(x))^{-2} \cdot \frac{d}{dx}(\cot(x) - \csc(x))$.
Step 3.2:
Use the sum rule to find the derivative of $\cot(x) - \csc(x)$: $-(\cot(x) - \csc(x))^{-2} \cdot \left(\frac{d}{dx}(\cot(x)) + \frac{d}{dx}(-\csc(x))\right)$.
Step 3.3:
The derivative of $\cot(x)$ is $-\csc^2(x)$: $-(\cot(x) - \csc(x))^{-2} \cdot \left(-\csc^2(x) + \frac{d}{dx}(-\csc(x))\right)$.
Step 3.4:
Differentiate $-\csc(x)$, noting that $-1$ is a constant: $-(\cot(x) - \csc(x))^{-2} \cdot \left(-\csc^2(x) - \frac{d}{dx}(\csc(x))\right)$.
Step 3.5:
The derivative of $\csc(x)$ is $-\csc(x)\cot(x)$: $-(\cot(x) - \csc(x))^{-2} \cdot \left(-\csc^2(x) - (-\csc(x)\cot(x))\right)$.
Step 3.6:
Perform the multiplication.
Step 3.6.1:
Multiply $-1$ by $-1$: $-(\cot(x) - \csc(x))^{-2} \cdot (\csc^2(x) - \csc(x)\cot(x))$.
Step 3.6.2:
Multiply $\csc(x)$ by $1$: $-(\cot(x) - \csc(x))^{-2} \cdot (\csc^2(x) - \csc(x)\cot(x))$.
Step 4:
Combine the left and right sides of the equation: $y = -(\cot(x) - \csc(x))^{-2} \cdot (\csc^2(x) - \csc(x)\cot(x))$.
Step 5:
Substitute $\frac{dy}{dx}$ for $y$: $\frac{dy}{dx} = -(\cot(x) - \csc(x))^{-2} \cdot (\csc^2(x) - \csc(x)\cot(x))$.
Knowledge Notes:
To solve this problem, we need to understand several calculus concepts and rules for differentiation:
Chain Rule: This rule is used when differentiating composite functions. If $f(x) = h(g(x))$, then the derivative of $f$ with respect to $x$ is $f'(x) = h'(g(x)) \cdot g'(x)$.
Power Rule: This rule states that if $f(x) = x^n$, where $n$ is any real number, then the derivative of $f$ with respect to $x$ is $f'(x) = nx^{n-1}$.
Sum Rule: The derivative of a sum of functions is the sum of the derivatives of those functions. If $f(x) = u(x) + v(x)$, then $f'(x) = u'(x) + v'(x)$.
Derivatives of Trigonometric Functions: The derivative of $\cot(x)$ is $-\csc^2(x)$, and the derivative of $\csc(x)$ is $-\csc(x)\cot(x)$.
Negative Constants: When differentiating, a negative constant multiplies through the derivative. If $f(x) = -g(x)$, then $f'(x) = -g'(x)$.
Understanding and applying these rules allows us to differentiate complex expressions involving trigonometric functions and their inverses.
