Find y' for y = 2^x + x²
Find $y'$ for $y = 2^{x} + x^{2}$.
Explanation of Method
Distinguish a power function from an exponential function.
Step 1: Identify the two different rules needed
The two terms look similar but are structurally different: in $2^{x}$ the variable sits in the exponent, while in $x^{2}$ it sits in the base. They therefore call for different rules:
\[ \bigl(a^{x}\bigr)' = a^{x}\ln a \quad\text{(Problem 18)}, \qquad \bigl(x^{n}\bigr)' = nx^{n-1} \quad\text{(Problem 14)}. \]
Step 2: Differentiate term by term
\[ \frac{d}{dx}\bigl(2^{x}\bigr) = 2^{x}\ln 2, \qquad \frac{d}{dx}\bigl(x^{2}\bigr) = 2x. \]
Step 3: Verification
\[x = 1:\quad y' = 2\ln 2 + 2 \approx 1.3863 + 2 = 3.3863\]
\[\frac{y(1+10^{-5})-y(1-10^{-5})}{2\times 10^{-5}} \approx 3.3863 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = 2^{x}\ln 2 + 2x\,}\]
Remark
Remark. Both terms are positive for $x \gt 0$, so the function increases there. The common error is to write $\bigl(2^{x}\bigr)' = x\,2^{x-1}$, applying the power rule to an exponential; the factor $\ln 2$ is the signature of the correct rule.