Question

If \[\operatorname{u}=f({{x}^{3}})\] , \[\operatorname{v}=g({{x}^{2}})\] , \[\operatorname{f}'(x)=cosx\] and \[\operatorname{g}'(x)=sinx\] , then find the value of \[\dfrac{du}{dv}\] .
A. \[\dfrac{3}{2}xcos{{x}^{3}}cosec{{x}^{2}}\]
B. \[\dfrac{2}{3}sin{{x}^{3}}sec{{x}^{2}}\]
C. \[\operatorname{tanx}\]
D. None of these

Answer 1

Hint: Integrate \[\operatorname{f}'(x)\] and \[\operatorname{g}'(x)\] with respect to \[dx\] . After integration, we get \[\operatorname{f}(x)=sinx\] and
\[\operatorname{g}(x)=-cosx\] . Now, we have \[\operatorname{u}=f({{x}^{3}})=sin{{x}^{3}}\] and \[\operatorname{v}=g({{x}^{2}})=-cos{{x}^{2}}\] . Using the chain rule, we can write \[\dfrac{du}{dv}\] as \[\dfrac{du}{dx}.\dfrac{dx}{dv}\] . Now, find \[\dfrac{du}{dx}\] and \[\dfrac{dx}{dv}\] . Put their values in \[\dfrac{du}{dx}.\dfrac{dx}{dv}\] and solve them further.

Complete step-by-step answer:
In the question, it is given that \[\operatorname{u}=f({{x}^{3}})\] , \[\operatorname{v}=g({{x}^{2}})\] , \[\operatorname{f}'(x)=cosx\] and \[\operatorname{g}'(x)=sinx\] .
We have \[\operatorname{f}'(x)=cosx\] . We can write \[\operatorname{f}'(x)\] as \[\dfrac{df}{dx}\] .
Now, we have \[\dfrac{df}{dx}=cosx\]………………(1)
Integrating equation (1), we get
\[\begin{align}
  & df=\cos x\,dx \\
 & \Rightarrow \int{df}=\int{\cos x\,dx} \\
\end{align}\]
We know that the integration of \[\cos x\] is \[\sin x\] .
\[\Rightarrow \operatorname{f}(x)=sinx+{{c}_{1}}\]………………(2)
Where \[{{\operatorname{c}}_{1}}\] is constant.
We also have \[g'(x)=\operatorname{sinx}\] . We can write \[g'(x)\] as \[\dfrac{dg}{dx}\] .
Now, we have \[\dfrac{dg}{dx}=sinx\]…………..(3)
Integrating equation (3), we get
\[\begin{align}
  & dg=\sin x\,dx \\
 & \Rightarrow \int{dg}=\int{\sin x\,dx} \\
\end{align}\]
We know that the integration of \[\sin x\] is \[-\cos x\] .
\[\Rightarrow g(x)=-\operatorname{cosx}+{{c}_{2}}\]………………(4)
Where \[{{\operatorname{c}}_{2}}\] is constant.
From equation (2) and equation (4), we have \[f(x)=sinx+{{c}_{1}}\] , \[g(x)=-\operatorname{cosx}+{{c}_{2}}\] .
Now, we have to find \[u\] and \[v\] .
According to question, it is given that \[\operatorname{u}=f({{x}^{3}})\] and \[\operatorname{v}=g({{x}^{2}})\]
We have, \[\operatorname{f}(x)=sinx+{{c}_{1}}\] . We have to find \[\operatorname{f}({{x}^{3}})\] .
Substituting x by \[{{\operatorname{x}}^{3}}\] , we get \[\operatorname{f}({{x}^{3}})=sin{{x}^{3}}+{{c}_{1}}\] ………………(5)
We have, \[\operatorname{g}(x)=-cosx+{{c}_{2}}\] . We have to find \[\operatorname{g}({{x}^{2}})\] .
 Substituting x by \[{{\operatorname{x}}^{2}}\] , we get \[\operatorname{g}({{x}^{2}})=-cos{{x}^{2}}+{{c}_{1}}\] …………………..(6)
We have to find the value of \[\dfrac{du}{dv}\] .
Using the chain rule, we can transform \[\dfrac{du}{dv}\] as \[\dfrac{du}{dv}=\dfrac{du}{dx}.\dfrac{dx}{dv}\] ……..(7)
Now, we need \[\dfrac{du}{dx}\] and \[\dfrac{dv}{dx}\] .
\[\operatorname{u}=f({{x}^{3}})=sin{{x}^{3}}+{{c}_{1}}\]
\[\dfrac{du}{dx}=cos{{x}^{3}}.3{{x}^{2}}=3{{x}^{2}}.cos{{x}^{3}}\]………….(8)
\[\operatorname{g}({{x}^{2}})=-cos{{x}^{2}}+{{c}_{1}}\]
\[\dfrac{dv}{dx}=sin{{x}^{2}}.2x=2x.sin{{x}^{2}}\]……………….(9)
Using equation (8) and equation (9), we can transform equation (7)
\[\begin{align}
  & \dfrac{du}{dv}=\dfrac{du}{dx}.\dfrac{dx}{dv}=(3{{x}^{2}}.cos{{x}^{3}})\times \dfrac{1}{(2x.sin{{x}^{2}})} \\
 & \dfrac{du}{dv}=\dfrac{3{{x}^{2}}.cos{{x}^{3}}}{2x.sin{{x}^{2}}}=\dfrac{3}{2}xcos{{x}^{3}}cosec{{x}^{2}} \\
\end{align}\]
So, the value of \[\dfrac{du}{dv}=\dfrac{3}{2}xcos{{x}^{3}}cosec{{x}^{2}}\] .
Therefore, option (A) is correct.

Note: In this question, one can try to find \[\dfrac{du}{dv}\] without using the chain rule. One can solve this derivative by transforming u in terms of v. This approach will increase the complexity of the solution. So, try to find the derivative using the chain rule. It will be easier to solve.