Question

Help with Differentiating an equation. Picture included, with answer- need it explained

Answer 1

so this would equal 7sqrt(x) = 7x^(1/2)e^x
g'(x) = 7x^(1/2)e^x + 7e^x(1/2x^(-1/2))

Answer 2

then it would be
7/2)x^(1/2)*e^x??????

how does ((1/2)x^(-1/2)) become 2x+1

Answer 3

\[(7e^x\sqrt x)'=7(e^x\sqrt x)'\]
right?, now let 7 outside, since it is constant, we put it in later.
Apply product rule, you have
\[(e^x)' \sqrt x + e^x (\sqrt x)'= e^x \sqrt x+e^x \dfrac{1}{2\sqrt x}\]
put 7 in, and use \(\sqrt x = x^{1/2}\), you have
\[7e^x x^{1/2}+\dfrac{7}{2}e^x x^{-1/2}\]
multiple and divide the first term by 2, you get
\[\dfrac{2*7e^x x^{1/2}}{2}+\dfrac{7}{2}e^x x^{-1/2}\]
Now, factor \(\dfrac{7}{2}e^x x^{-1/2} out\)
Note: \(x^{1/2}= x^{2/2-1/2}= x^{2/2}*x^{-1/2}=x*x^{-1/2}\)
the answer follows

Answer 4

Thank you