Question Number 11150 by suci last updated on 14/Mar/17
$${f}\left({x}\right)=\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{{sin}\mathrm{2}{x}} \\ $$$${f}'\left({x}\right)=…??? \\ $$
Answered by ajfour last updated on 14/Mar/17
$$=\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{sin}\:\mathrm{2}{x}} \:\left\{\left(\mathrm{2cos}\:\mathrm{2}{x}\right)\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}\mathrm{sin}\:\mathrm{2}{x}\:\right\} \\ $$
Commented by ajfour last updated on 14/Mar/17
![taking log and differentiate : y=g(x)^(h(x)) ⇒ ln y =h(x)ln g(x) (1/y)(dy/dx) =h′(x)ln g(x)+h(x)g′(x)/g(x) (dy/dx)=g(x)^(h(x)) [h′(x)ln g(x)+h(x)((g′(x))/(g(x))) ]](https://www.tinkutara.com/question/Q11161.png)
$${taking}\:{log}\:{and}\:{differentiate}\:: \\ $$$${y}={g}\left({x}\right)^{{h}\left({x}\right)} \:\:\Rightarrow\:\mathrm{ln}\:{y}\:={h}\left({x}\right)\mathrm{ln}\:{g}\left({x}\right) \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}\:={h}'\left({x}\right)\mathrm{ln}\:{g}\left({x}\right)+{h}\left({x}\right){g}'\left({x}\right)/{g}\left({x}\right) \\ $$$$\frac{{dy}}{{dx}}={g}\left({x}\right)^{{h}\left({x}\right)} \left[{h}'\left({x}\right)\mathrm{ln}\:{g}\left({x}\right)+{h}\left({x}\right)\frac{{g}'\left({x}\right)}{{g}\left({x}\right)}\:\right] \\ $$