Let \( y=\tan (2 x+5) \). Find the differential \( d y \) when \( x=1 \) and \( d x=0.1 \) Find the differential \( d y \) when \( x=1 \) and \( d x=0.2 \)

To find the differential \( d y \) when \( y = \tan(2x + 5) \), we first need to determine the derivative \( \frac{dy}{dx} \). Using the chain rule: \[ \frac{dy}{dx} = \sec^2(2x + 5) \cdot \frac{d(2x + 5)}{dx} = \sec^2(2x + 5) \cdot 2 = 2 \sec^2(2x + 5). \] Next, we need to evaluate this at \( x = 1 \): \[ 2x + 5 = 2(1) + 5 = 7 \quad \Rightarrow \quad \sec^2(7) = \frac{1}{\cos^2(7)}. \] Now, we calculate \( d y \): \[ d y = \frac{dy}{dx} \cdot d x. \] ### For \( d x = 0.1 \): First, we calculate: \[ \frac{dy}{dx} \bigg|_{x=1} = 2 \sec^2(7), \] thus: \[ d y = 2 \sec^2(7) \cdot 0.1 = 0.2 \sec^2(7). \] ### For \( d x = 0.2 \): Similarly: \[ d y = 2 \sec^2(7) \cdot 0.2 = 0.4 \sec^2(7). \] In summary: - When \( x = 1 \) and \( d x = 0.1 \), \( d y = 0.2 \sec^2(7) \). - When \( x = 1 \) and \( d x = 0.2 \), \( d y = 0.4 \sec^2(7) \).