Question Number 44190 by olj55336@awsoo.com last updated on 23/Sep/18
$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${Q}..{Find}\:{second}\:{derivative}.. \\ $$$$\:\:\mathrm{5}^{{x}\:\:\:\:} {solve}\:{please}\:{stap}\:{by}\:{stap} \\ $$
Commented by maxmathsup by imad last updated on 23/Sep/18
$${let}\:{f}\left({x}\right)\:={a}^{{x}} \:\:\:{with}\:{a}>\mathrm{0}\:{and}\:{a}\neq\mathrm{1}\:{let}\:{find}\:{generally}\:\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$${we}\:{have}\:{f}\left({x}\right)\:={e}^{{xln}\left({a}\right)} \:\Rightarrow{f}^{\left(\mathrm{1}\right)} \left({x}\right)={ln}\left({a}\right)\:{e}^{{xln}\left({a}\right)} \:\Rightarrow{f}^{\left(\mathrm{2}\right)} \left({x}\right)=\left\{{ln}\left({a}\right)\right\}^{\mathrm{2}} \:{e}^{{xln}\left({a}\right)} \\ $$$${let}\:{suppose}\:{f}^{\left({n}\right)} \left({x}\right)=\left\{{ln}\left({a}\right)\right\}^{{n}} \:{e}^{{xln}\left({a}\right)} \:\Rightarrow{f}^{\left({n}+\mathrm{1}\right)} \left({x}\right)=\left\{{ln}\left({a}\right)\right\}^{{n}+\mathrm{1}} \:{e}^{{xln}\left({a}\right)} \\ $$$${so}\:{f}^{\left({n}\right)} \left({x}\right)=\left\{{ln}\left({a}\right)\right\}^{{n}} \:{a}^{{x}} \\ $$$${for}\:{f}\left({x}\right)=\mathrm{5}^{{x}} \:\Rightarrow{f}^{\left(\mathrm{2}\right)} \left({x}\right)=\left\{{ln}\left(\mathrm{5}\right)\right\}^{\mathrm{2}} \:\mathrm{5}^{{x}} \:. \\ $$
Answered by MrW3 last updated on 23/Sep/18
$${y}=\mathrm{5}^{{x}} ={e}^{{x}\mathrm{ln}\:\mathrm{5}} \\ $$$$\frac{{dy}}{{dx}}=\left(\mathrm{ln}\:\mathrm{5}\right){e}^{{x}\mathrm{ln}\:\mathrm{5}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\left(\mathrm{ln}\:\mathrm{5}\right)^{\mathrm{2}} {e}^{{x}\mathrm{ln}\:\mathrm{5}} =\left(\mathrm{ln}\:\mathrm{5}\right)^{\mathrm{2}} \mathrm{5}^{{x}} \\ $$