Find dy for y = 2^x-3^-x+ sqrt(x), with x>0
Find $dy$ for $y = 2^{x}-3^{-x}+\sqrt{x}$, with $x>0$.
Explanation of Method
Three independent terms, and a sign that is easy to lose.
Step 1: The first term
The rule for a general exponential base is $\bigl(a^{x}\bigr)' = a^{x}\ln a$: $\frac{d}{dx}\bigl(2^{x}\bigr) = 2^{x}\ln 2$.
Step 2: The second term — two minus signs
Here the exponent is $-x$, so the chain rule contributes a factor $-1$, and the term itself already carries a minus sign. The two cancel: $\frac{d}{dx}\bigl(-3^{-x}\bigr) = -3^{-x}\ln 3\cdot(-1) = +\,3^{-x}\ln 3$.
This is the one step in the problem where a sign is commonly dropped.
Step 3: The third term
\[ \frac{d}{dx}\bigl(\sqrt{x}\bigr) = \frac{1}{2\sqrt{x}}. \]
Step 4: Assemble and form the differential
\[ y' = 2^{x}\ln 2 + 3^{-x}\ln 3 + \frac{1}{2\sqrt{x}}, \qquad dy = y'\,dx. \]
Step 5: Verification
\[x = 1:\quad y' = 2\ln 2 + \tfrac13\ln 3 + \tfrac12 \approx 1.386294+0.366204+0.5 = 2.252498\]
\[\frac{y(1.0001)-y(0.9999)}{0.0002} \approx 2.252498 \;\;\checkmark\]
\[\text{All three pieces are positive for } x>0, \text{ so } y \text{ is strictly increasing} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,dy = \left(2^{x}\ln 2 + 3^{-x}\ln 3 + \dfrac{1}{2\sqrt{x}}\right)dx\,}\]
Remark
Remark. Note that $3^{-x}$ could equally be written $\left(\tfrac13\right)^{x}$, and then the rule gives $\left(\tfrac13\right)^{x}\ln\tfrac13 = -3^{-x}\ln 3$ directly — with the leading minus of the term this again produces $+3^{-x}\ln 3$. The two routes agree, which is a useful check on the sign.