Question

Find the derivative of the following function
f(x) = $\frac{3^x\cot 45°}{2x}$

• A. $f\text{'}\left(x\right)=\frac{3^x\times \left(2x\right)+2\left(3^x\right)}{2x^2}$
• B. $f\text{'}\left(x\right)=\frac{3^x\times \left(2x\right)+2\left(3^x\right)}{4x^2}$
• C. $f\text{'}\left(x\right)=\frac{3^x\ln 3\times \left(2x\right)-2\left(3^x\right)}{4x^2}$
• D. $f\text{'}\left(x\right)=\frac{3^x\times \left(2x\right)-2\left(3^x\right)}{4x^2}$

Answer 1

The correct option is C $f\text{'}\left(x\right)=\frac{3^x\ln 3\times \left(2x\right)-2\left(3^x\right)}{4x^2}$

$\cot 45°=\mathrm{constant}=1f\left(x\right)=\frac{3^x}{2x}$
Using division rule -
$f\text{'}\left(x\right)=\frac{\frac{d\left(3^x\right)}{\mathrm{dx}}\times \left(2x\right)-\frac{d\left(2x\right)}{\mathrm{dx}}\times 3^x}{{\left(2x\right)}^2}$

$f\text{'}\left(x\right)=\frac{3^x\ln 3\times \left(2x\right)-2\left(3^x\right)}{4x^2}$