Looking to solve :
$$ \frac{d}{dx}[2^{0.5x}]$$
The multiplication and X value in the exponent is confusing me. Help? Thanks!
$\endgroup$ 34 Answers
$\begingroup$It might help you to rewrite $2^{0.5x}$ as $2^{0.5 x} = e^{ln(2^{0.5x})} = e^{0.5x \cdot ln(2)}$. Are you familiar with why this re-writing works? Can you think of how to derive it now in the last way I wrote it?
If you still have problems, let me know and I can answer in more detail.
$\endgroup$ 6 $\begingroup$This is something called logarithmic differentiation, the technique as I learned it goes like this, and makes use of implicit differentiation and thinking of $d$ as an operator which can be applied to both sides of an equation. Setting: $$y= 2^{.5x}$$ We have the relationship between $x$ and $y$ (think implicit function material) $$\ln(y)=.5x\ln(2)$$ Then, applying the $d$ operator $$d[\ln(y)]=d[.5x\ln(2)]\Rightarrow \frac{dy}{y}=.5\ln(2)dx $$ As requested, we have the LHS because $$d[\ln(y)]=\frac{1}{y}*dy$$ by the chain rule, first taking the derivative of $ln$ and then multiplying by the derivative of the inside, here $y$, which is $dy$. Might be worth reviewing implicit differentiation.
Then solving for $\frac{dy}{dx}$ we have $$\frac{dy}{dx}=.5\ln(2)y=.5\ln(2)2^{.5x} $$
Which is also the general derivative of $c^{ax}$. Give it a shot.
$\endgroup$ 2 $\begingroup$$$\frac { d }{ dx } \left( { 2 }^{ 0.5x } \right) ={ 2 }^{ 0.5x }\ln { 2 } { \left( 0.5x \right) }^{ \prime }={ 2 }^{ 0.5x }\ln { \sqrt { 2 } } $$
$\endgroup$ $\begingroup$You have to write
$$2^{0.5x}=e^{0.5x\log(2)},$$
so
$$\frac d{dx}\left(e^{0.5x\log(2)}\right)=0.5\log(2)e^{0.5x\log(2)}.$$
Finally,
$$\frac d{dx}(2^{0.5x})=\frac{\log(2)}2 2^{0.5x}.$$
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