Question

find dy/dx if y = (cscx + 1)^7

Explanation:

Step1: Identify the outer and inner functions

Let \( u = \csc x + 1 \), so \( y = u^7 \). We use the chain rule, which states that \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \).

Step2: Differentiate the outer function

Differentiate \( y = u^7 \) with respect to \( u \). Using the power rule \( \frac{d}{du}(u^n)=nu^{n - 1} \), we get \( \frac{dy}{du}=7u^{6} \). Substituting back \( u=\csc x + 1 \), we have \( \frac{dy}{du}=7(\csc x + 1)^{6} \).

Step3: Differentiate the inner function

Differentiate \( u=\csc x + 1 \) with respect to \( x \). The derivative of \( \csc x \) is \( -\csc x\cot x \) and the derivative of a constant (1) is 0. So \( \frac{du}{dx}=-\csc x\cot x \).

Step4: Apply the chain rule

Multiply \( \frac{dy}{du} \) and \( \frac{du}{dx} \): \( \frac{dy}{dx}=7(\csc x + 1)^{6}\cdot(-\csc x\cot x)=- 7(\csc x + 1)^{6}(\csc x\cot x) \).

Answer:

\(-7(\csc x + 1)^{6}(\csc x\cot x)\) (corresponding to the orange - colored option: \(-7(\csc x + 1)^{6}(\csc x\cot x)\))