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 \( dy \), we first need to find the derivative of \( y \) with respect to \( x \). Given: \[ y = \tan(2x + 5) \] To find \( \frac{dy}{dx} \), we can use the chain rule: \[ \frac{dy}{dx} = \sec^2(2x + 5) \cdot \frac{d}{dx}(2x + 5) = \sec^2(2x + 5) \cdot 2 = 2\sec^2(2x + 5) \] Now, we will find \( dy \): \[ dy = \frac{dy}{dx} \cdot dx \] ### When \( x = 1 \) and \( dx = 0.1 \): First, we calculate \( 2x + 5 \) at \( x = 1 \): \[ 2(1) + 5 = 7 \] Now we find \( \sec^2(7) \): \[ \frac{dy}{dx}\bigg|_{x=1} = 2\sec^2(7) \] Then for \( dx = 0.1 \): \[ dy = 2\sec^2(7) \cdot 0.1 \] So when \( x=1 \) and \( dx=0.1 \): \[ dy \approx 0.2\sec^2(7) \] ### When \( x = 1 \) and \( dx = 0.2 \): Using the same \( \sec^2(7) \): \[ dy = 2\sec^2(7) \cdot 0.2 \] So when \( x=1 \) and \( dx=0.2 \): \[ dy \approx 0.4\sec^2(7) \] Finally, substituting the approximate value of \( \sec^2(7) \) for numerical results, we keep in mind the calculations would be completed using a calculator for final numerical values.