Question
If $$u = f\left( {{x^3}} \right),\,v = g\left( {{x^2}} \right),\,f'\left( x \right) = \cos \,x$$ and $$g'\left( x \right) = \sin \,x,$$ then $$\frac{{du}}{{dv}} = ?$$
A.
$$\frac{1}{2}x\,\cos \,{x^3}{\text{cosec }}{x^2}$$
B.
$$\frac{3}{2}x\,\cos \,{x^3}{\text{cosec }}{x^2}$$
C.
$$\frac{1}{2}x\,\sec \,{x^3}{\text{sin }}{x^2}$$
D.
$$\frac{3}{2}x\,\sec \,{x^3}{\text{cosec }}{x^2}$$
Answer :
$$\frac{3}{2}x\,\cos \,{x^3}{\text{cosec }}{x^2}$$
Solution :
$$\eqalign{
& {\text{Here, }}u = f\left( {{x^3}} \right) \cr
& \Rightarrow \frac{{du}}{{dx}} = f'\left( {{x^3}} \right).\frac{d}{{dx}}\left( {{x^3}} \right) \cr
& = \left( {\cos \left( {{x^3}} \right)} \right).3{x^2} \cr
& = 3{x^2}.\cos \,{x^3} \cr
& {\text{and }}v = g\left( {{x^2}} \right) \cr
& \Rightarrow \frac{{dv}}{{dx}} = g'\left( {{x^2}} \right).\frac{d}{{dx}}\left( {{x^2}} \right) \cr
& = \left( {\sin \,{x^2}} \right).\left( {2x} \right) \cr
& = 2x.\sin \,{x^2} \cr
& \therefore \,\frac{{du}}{{dv}} = \frac{{\frac{{du}}{{dx}}}}{{\frac{{dv}}{{dx}}}}\frac{{3{x^2}.\cos \,{x^3}}}{{2x.\sin \,{x^2}}} \cr
& \Rightarrow \frac{{du}}{{dv}} = \frac{3}{2}x.\cos \,{x^3}.{\text{cosec}}\,{x^2} \cr} $$