Calculate the value of the differential of y =

Calculate the value of the differential of $y = 3^{1/x}+2^{-2x}+6^{\sqrt{x}}$ at $x = 1$ with $dx = 0.2$.

Step 1: Differentiate the first term

With $u = \dfrac1x$, so $u' = -\dfrac{1}{x^{2}}$: $\frac{d}{dx}\bigl(3^{1/x}\bigr) = 3^{1/x}\ln 3\cdot\left(-\frac{1}{x^{2}}\right)$.

Step 2: Differentiate the second term

With $v = -2x$, so $v' = -2$: $\frac{d}{dx}\bigl(2^{-2x}\bigr) = 2^{-2x}\ln 2\cdot(-2) = -2\ln 2\cdot 2^{-2x}$.

Step 3: Differentiate the third term

With $w = \sqrt{x}$, so $w' = \dfrac{1}{2\sqrt{x}}$: $\frac{d}{dx}\bigl(6^{\sqrt{x}}\bigr) = 6^{\sqrt{x}}\ln 6\cdot\frac{1}{2\sqrt{x}}$.

Step 4: Substitute $x = 1$

There $3^{1/x} = 3$, $2^{-2x} = \dfrac14$, $6^{\sqrt{x}} = 6$, and all three inner derivatives are easy: \[y'(1) = 3\ln 3\cdot(-1) \;-\; 2\ln 2\cdot\frac14 \;+\; 6\ln 6\cdot\frac12 = -3\ln 3-\frac{\ln 2}{2}+3\ln 6.\]

Step 5: The logarithms collapse

Since $\ln 6 = \ln 2+\ln 3$, the terms in $\ln 3$ cancel exactly: \[y'(1) = -3\ln 3-\frac{\ln 2}{2}+3\ln 2+3\ln 3 = 3\ln 2-\frac{\ln 2}{2} = \frac{5\ln 2}{2} \approx 1.7328680.\]

Step 6: Form the differential

\[ dy = y'(1)\,dx = \frac{5\ln 2}{2}\cdot 0.2 = \frac{\ln 2}{2}. \]

Step 7: Verification

\[y(1) = 3+0.25+6 = 9.25\]

\[\frac{y(1.0001)-y(0.9999)}{0.0002} \approx 1.732868 = \frac{5\ln 2}{2} \;\;\checkmark\]

\[\text{True increment over } \Delta x = 0.2:\ y(1.2)-y(1) \approx 0.55656 \text{ versus } dy \approx 0.34657\]