Question Number 187916 by mustafazaheen last updated on 23/Feb/23
$${how}\:{is}\:{solution} \\ $$$${y}=\left({cosx}\right)^{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } } \\ $$$$\frac{{dy}}{{dx}}=? \\ $$
Answered by mr W last updated on 23/Feb/23
$${generally} \\ $$$${y}={f}\left({x}\right)^{{g}\left({x}\right)} ={e}^{{g}\left({x}\right)\mathrm{ln}\:{f}\left({x}\right)} \\ $$$${y}'={f}\left({x}\right)^{{g}\left({x}\right)} \left[{g}'\left({x}\right)\mathrm{ln}\:{f}\left({x}\right)+\frac{{g}\left({x}\right){f}'\left({x}\right)}{{f}\left({x}\right)}\right] \\ $$$$ \\ $$$${y}=\left(\mathrm{cos}\:{x}\right)^{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } } \\ $$$${y}'=\left(\mathrm{cos}\:{x}\right)^{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } } \left\{\left[\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } \right]^{,} \mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)−\frac{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } \mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\right\} \\ $$$$\:\:=\left(\mathrm{cos}\:{x}\right)^{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } } \left\{\left[\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } \left({e}^{{x}} \mathrm{ln}\:\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)+\frac{\mathrm{6}{xe}^{{x}} }{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}}\right)\right]\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)−\frac{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } \mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\right\} \\ $$$$\:\:=\left(\mathrm{cos}\:{x}\right)^{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } } \left\{\left[\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } {e}^{{x}} \left(\mathrm{ln}\:\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)+\frac{\mathrm{6}{x}}{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}}\right)\right]\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)−\frac{\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)^{{e}^{{x}} } \mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\right\} \\ $$
Commented by mustafazaheen last updated on 23/Feb/23
$${Thanks}\:{Mr} \\ $$