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Prove the following expression if it is given that ${{x}^{16}}{{y}^{9}}={{\left( {{x}^{2}}+y \right)}^{17}}$.
 $\dfrac{dy}{dx}=\dfrac{2y}{x}$.

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Hint: Here, we will use differentiation to prove the given statement that $\dfrac{dy}{dx}=\dfrac{2y}{x}$. We will differentiate both side of thee given equation that is ${{x}^{16}}{{y}^{9}}={{\left( {{x}^{2}}+y \right)}^{17}}$ with respect to x and hence find the value of $\dfrac{dy}{dx}$. We can use product rule of differentiation here which is given as, if u and v are two functions of x then the differentiation of product of u and v is given as $\dfrac{d\left( u.v \right)}{dx}=u.\dfrac{dv}{dx}+v.\dfrac{du}{dx}$.

Complete step-by-step answer:
Since, the equation given to us is:
${{x}^{16}}{{y}^{9}}={{\left( {{x}^{2}}+y \right)}^{17}}$
On differentiating both sides of this equation with respect to x, we get:
${{x}^{16}}9{{y}^{8}}\dfrac{dy}{dx}+{{y}^{9}}16{{x}^{15}}=17{{\left( {{x}^{2}}+y \right)}^{16}}.\left( 2x+\dfrac{dy}{dx} \right)$
On dividing and multiplying the first term by y and the second term by x (terms of left hand side) we can also write it as:
\[\dfrac{16{{x}^{16}}{{y}^{9}}}{x}+\dfrac{9{{x}^{16}}{{y}^{9}}}{y}\dfrac{dy}{dx}=17{{\left( {{x}^{2}}+y \right)}^{16}}.\left( 2x+\dfrac{dy}{dx} \right)\]
On taking ${{x}^{16}}{{y}^{9}}$ common on LHS, we can write as:
${{x}^{16}}{{y}^{9}}\left( \dfrac{16}{x}+\dfrac{9}{y}\dfrac{dy}{dx} \right)=17{{\left( {{x}^{2}}+y \right)}^{16}}\left( 2x+\dfrac{dy}{dx} \right)$
Since, ${{x}^{16}}{{y}^{9}}={{\left( {{x}^{2}}+y \right)}^{17}}$, using it we can write:
${{\left( {{x}^{2}}+y \right)}^{17}}\left( \dfrac{16}{x}+\dfrac{9}{y}.\dfrac{dy}{dx} \right)=17{{\left( {{x}^{2}}+y \right)}^{16}}\left( 2x+\dfrac{dy}{dx} \right)$
On cancelling ${{\left( {{x}^{2}}+y \right)}^{16}}$ from both sides, we get:
$\left( {{x}^{2}}+y \right)\left( \dfrac{16}{x}+\dfrac{9}{y}\dfrac{dy}{dx} \right)=17\left( 2x+\dfrac{dy}{dx} \right)$
On multiplying by opening the brackets, we get:
   \[\begin{align}
  & {{x}^{2}}\dfrac{16}{x}+{{x}^{2}}.\dfrac{9}{y}.\dfrac{dy}{dx}+\dfrac{16y}{x}+\dfrac{9}{y}.y\dfrac{dy}{dx}=34x+17\dfrac{dy}{dx} \\
 & \Rightarrow 16x+\left( \dfrac{9{{x}^{2}}}{y} \right)\dfrac{dy}{dx}+\dfrac{16y}{x}+9\dfrac{dy}{dx}=34x+17\dfrac{dy}{dx} \\
 & \Rightarrow \dfrac{9{{x}^{2}}}{y}\dfrac{dy}{dx}+\dfrac{16y}{x}=34x-16x+17\dfrac{dy}{dx}-9\dfrac{dy}{dx} \\
 & \Rightarrow \dfrac{9{{x}^{2}}}{y}\dfrac{dy}{dx}-8\dfrac{dy}{dx}=18x-\dfrac{16y}{x} \\
 & \Rightarrow \left( \dfrac{9{{x}^{2}}}{y}-8 \right)\dfrac{dy}{dx}=\dfrac{2y}{x}\left( \dfrac{9{{x}^{2}}}{y}-8 \right) \\
\end{align}\]
On cancelling $\left( \dfrac{9{{x}^{2}}}{y}-8 \right)$ from both sides, we get:
$\dfrac{dy}{dx}=\dfrac{2y}{x}$
Hence, we have proved the given expression using the given equation.

Note: Students should note here that while differentiating the left hand side of the given equation, we have used the product rule which states that $\dfrac{d\left( u.v \right)}{dx}=u.\dfrac{dv}{dx}+v.\dfrac{du}{dx}$, where u and v are functions of x.
Students should observe in the last step that we can write $18x-\dfrac{16y}{x}$ as $\dfrac{2y}{x}\left( \dfrac{9{{x}^{2}}}{y}-8 \right)$, so that the term $\left( \dfrac{9{{x}^{2}}}{y}-8 \right)$ can be cancelled from the both sides.