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

& df=\cos x\,dx \\\ & \Rightarrow \int{df}=\int{\cos x\,dx} \\\ \end{aligned}$$ We know that the integration of $$\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{aligned} & dg=\sin x\,dx \\\ & \Rightarrow \int{dg}=\int{\sin x\,dx} \\\ \end{aligned}$$ We know that the integration of $$\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{aligned} & \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{aligned}$$ So, the value of $$\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.