Here's the question you clicked on:
101Ryan101
differentiate 3^(-x/2)
y=3^(-x/2) logy=(-x/2)log3 1/y(dy/dx)=-(1/2)log3 dy/dx=y*(-1/2)log3 dy/dx=(3^(-x/2))(-1/2)log3
should be (3^(-x/2))(-x/2)log3
d/dx(Cx^Ax)=Cx^Ax(Aln(Cx)+AxC/Cx)
d/dx(C^Ax)=C^Ax(Aln(C))
\[f(x)=3^{\frac{x}{2}}=e^{\frac{\ln(3) \times x}{2}}\] \[f'(x)=\frac{\ln(3)}{2} \times e^{\frac{\ln(3) \times x}{2}}=\frac{\ln(3)}{2} \times 3^{\frac{x}{2}}\]
Umm sorry, forgot the minus symbol, just plug it in and you get \[f'(x)=- \frac{\ln(3)}{2} \times 3^{- \frac{x}{2}} \]
no, the derivative of C^u is (C^u)(du)(lnC). So, what I did originally, but with ln instead of log. Someone1348's right, too