Hello, I am stuck on differentiating this:
6 * e^(x^(1/2))
Can someone please show me the steps. I understand chain and product rules will be implemented but I cannot seem to get it working.
Thanks kindly for any help.
Hello, I am stuck on differentiating this:
6 * e^(x^(1/2))
Can someone please show me the steps. I understand chain and product rules will be implemented but I cannot seem to get it working.
Thanks kindly for any help.
$\displaystyle \displaystyle \begin{align*} y = 6e^{x^{\frac{1}{2}}} \end{align*} $
Let $\displaystyle \displaystyle \begin{align*} u = x^{\frac{1}{2}} \end{align*} $ so that $\displaystyle \displaystyle \begin{align*} y = 6e^u \end{align*} $, then
$\displaystyle \displaystyle \begin{align*} \frac{du}{dx} = \frac{1}{2}x^{-\frac{1}{2}} \end{align*} $
$\displaystyle \displaystyle \begin{align*} \frac{dy}{du} &= 6e^u \\ &= 6e^{x^{\frac{1}{2}}} \end{align*} $
So
$\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} &= \frac{du}{dx} \cdot \frac{dy}{du} \\ &= \frac{1}{2}x^{-\frac{1}{2}} \cdot 6e^{x^{\frac{1}{2}}} \\ &= 3x^{-\frac{1}{2}}e^{x^{\frac{1}{2}}} \end{align*} $
$\displaystyle y' = {\left( 6 \right)^'} \cdot {{\rm{e}}^{\sqrt x }} + 6 \cdot {\left( {{{\rm{e}}^{\sqrt x }}} \right)^'} = 6 \cdot {\left( {{{\rm{e}}^{\sqrt x }}} \right)^'} = 6 \cdot {{\rm{e}}^{\sqrt x }} \cdot {\left( {\sqrt x } \right)^'} = \frac{3}{{\sqrt x }} \cdot {{\rm{e}}^{\sqrt x }}$