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Question Number 72250 by aliesam last updated on 26/Oct/19

f(x)=(ax+b)sin(x)+(cx+d)cos(x)  and   (dy/dx)=xcos(x)  find  a , b , c , d

$${f}\left({x}\right)=\left({ax}+{b}\right){sin}\left({x}\right)+\left({cx}+{d}\right){cos}\left({x}\right) \\ $$$${and}\:\:\:\frac{{dy}}{{dx}}={xcos}\left({x}\right) \\ $$$${find}\:\:{a}\:,\:{b}\:,\:{c}\:,\:{d} \\ $$$$ \\ $$$$ \\ $$

Commented by kaivan.ahmadi last updated on 27/Oct/19

(dy/dx)=asinx+(ax+b)cosx+c.cosx−(cx+d)sinx=  (a−cx−d)sinx+(ax+b+c)cosx⇒   { ((a−cx−d=0)),((ax+b+c=x⇒a=1,b+c=0⇒b=−c)) :}  ⇒1−cx−d=0⇒cx+d=1⇒c=0,d=1  ⇒b=0  ⇒a=d=1,b=c=0

$$\frac{{dy}}{{dx}}={asinx}+\left({ax}+{b}\right){cosx}+{c}.{cosx}−\left({cx}+{d}\right){sinx}= \\ $$$$\left({a}−{cx}−{d}\right){sinx}+\left({ax}+{b}+{c}\right){cosx}\Rightarrow \\ $$$$\begin{cases}{{a}−{cx}−{d}=\mathrm{0}}\\{{ax}+{b}+{c}={x}\Rightarrow{a}=\mathrm{1},{b}+{c}=\mathrm{0}\Rightarrow{b}=−{c}}\end{cases} \\ $$$$\Rightarrow\mathrm{1}−{cx}−{d}=\mathrm{0}\Rightarrow{cx}+{d}=\mathrm{1}\Rightarrow{c}=\mathrm{0},{d}=\mathrm{1} \\ $$$$\Rightarrow{b}=\mathrm{0} \\ $$$$\Rightarrow{a}={d}=\mathrm{1},{b}={c}=\mathrm{0} \\ $$$$ \\ $$

Commented by kaivan.ahmadi last updated on 27/Oct/19

f(x)=xsinx+cosx

$${f}\left({x}\right)={xsinx}+{cosx} \\ $$$$ \\ $$

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