Question

If a function is given as\[y=\dfrac{1}{a+\sqrt{x}}\], then find the value of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\].

Answer 1

Hint: The first order derivative is found using the formula $\dfrac{d}{dx}({{u}^{n}})=n{{u}^{n-1}}\dfrac{d}{dx}(u)$. And the second order derivative is found using the formula \[\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\].

Complete step-by-step solution -
The given expression is
\[y=\dfrac{1}{a+\sqrt{x}}\]
This can be re-written as,
\[y={{\left( a+\sqrt{x} \right)}^{-1}}\]
Now we will find the first order derivative of the given expression, so we will differentiate the given expression with respect to $'x'$, we get
\[\dfrac{d}{dx}(y)=\dfrac{d}{dx}\left( {{\left( a+\sqrt{x} \right)}^{-1}} \right)\]
Now we know $\dfrac{d}{dx}({{u}^{n}})=n{{u}^{n-1}}\dfrac{d}{dx}(u)$ , applying this formula, the above equation becomes,
\[\dfrac{dy}{dx}=(-1){{\left( a+\sqrt{x} \right)}^{-1-1}}\dfrac{d}{dx}\left( a+\sqrt{x} \right)\]
Now we will apply the the sum rule of differentiation, i.e., differentiation of sum of two functions is same as the sum of individual differentiation of the functions, i.e., $\dfrac{d}{dx}(u+v)=\dfrac{d}{x}(u)+\dfrac{d}{x}(v)$ . Applying this formula in the above equation, we get
\[\dfrac{dy}{dx}=(-1){{\left( a+\sqrt{x} \right)}^{-2}}\left[ \dfrac{d}{dx}\left( a \right)+\dfrac{d}{dx}{{\left( x \right)}^{\dfrac{1}{2}}} \right]\]
We know the differentiation of constant term is always zero, so
\[\dfrac{dy}{dx}=(-1){{\left( a+\sqrt{x} \right)}^{-2}}\left[ 0+\dfrac{d}{dx}{{\left( x \right)}^{\dfrac{1}{2}}} \right]\]
Now applying the formula $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$, the above equation becomes,
\[\dfrac{dy}{dx}=\dfrac{-1}{{{\left( a+\sqrt{x} \right)}^{2}}}\times \dfrac{1}{2}\times {{\left( x \right)}^{\dfrac{1}{2}-1}}\]
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{-1}{{{\left( a+\sqrt{x} \right)}^{2}}}\times \dfrac{1}{2}\times {{\left( x \right)}^{\dfrac{1-2}{2}}} \\
 & \dfrac{dy}{dx}=\dfrac{-1}{{{\left( a+\sqrt{x} \right)}^{2}}}\times \dfrac{1}{2}\times {{\left( x \right)}^{\dfrac{-1}{2}}} \\
 & \dfrac{dy}{dx}=\dfrac{-1}{2}{{\left( a+\sqrt{x} \right)}^{-2}}\times {{\left( x \right)}^{\dfrac{-1}{2}}} \\
\end{align}\]
Now we need to find the second order derivative, so we will differentiate the above equation with respect to $'x'$, we get
\[\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( \dfrac{-1}{2}{{\left( a+\sqrt{x} \right)}^{-2}}\times {{\left( x \right)}^{\dfrac{-1}{2}}} \right)\]
Taking out the constant term, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{2}\dfrac{d}{dx}\left( {{\left( a+\sqrt{x} \right)}^{-2}}\times {{\left( x \right)}^{\dfrac{-1}{2}}} \right)\]
We know the product rule as, \[\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\], applying this formula in the above equation, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{2}\left[ {{\left( a+\sqrt{x} \right)}^{-2}}\dfrac{d}{dx}\left( {{\left( x \right)}^{\dfrac{-1}{2}}} \right)+{{\left( x \right)}^{\dfrac{-1}{2}}}\dfrac{d}{dx}\left( {{\left( a+\sqrt{x} \right)}^{-2}} \right) \right]\]
Now we know $\dfrac{d}{dx}({{u}^{n}})=n{{u}^{n-1}}\dfrac{d}{dx}(u)$ , applying this formula, the above equation becomes,
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{2}\left[ {{\left( a+\sqrt{x} \right)}^{-2}}\left( \dfrac{-1}{2} \right){{\left( x \right)}^{\dfrac{-1}{2}-1}}+{{\left( x \right)}^{\dfrac{-1}{2}}}(-2){{\left( a+\sqrt{x} \right)}^{-2-1}}\dfrac{d}{dx}\left( a+\sqrt{x} \right) \right]\]
Now we will apply the the sum rule of differentiation, i.e., differentiation of sum of two functions is same as the sum of individual differentiation of the functions, i.e., $\dfrac{d}{dx}(u+v)=\dfrac{d}{x}(u)+\dfrac{d}{x}(v)$ . Applying this formula in the above equation, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{2}\left[ {{\left( a+\sqrt{x} \right)}^{-2}}\left( \dfrac{-1}{2} \right){{\left( x \right)}^{\dfrac{-1-2}{2}}}-2{{\left( x \right)}^{\dfrac{-1}{2}}}{{\left( a+\sqrt{x} \right)}^{-3}}\left[ \dfrac{d}{dx}\left( a \right)+\dfrac{d}{dx}{{\left( x \right)}^{\dfrac{1}{2}}} \right] \right]\]
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{2}\left[ \dfrac{1}{{{\left( a+\sqrt{x} \right)}^{2}}}\left( \dfrac{-1}{2} \right){{\left( x \right)}^{\dfrac{-3}{2}}}-\dfrac{2}{\sqrt{x}}.\dfrac{1}{{{\left( a+\sqrt{x} \right)}^{3}}}\left[ 0+\dfrac{1}{2}{{\left( x \right)}^{\dfrac{1}{2}-1}} \right] \right] \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{2}\left[ \dfrac{1}{{{\left( a+\sqrt{x} \right)}^{2}}}\left( \dfrac{-1}{2} \right)\dfrac{1}{{{\left( x \right)}^{\dfrac{3}{2}}}}-\dfrac{2}{\sqrt{x}}.\dfrac{1}{{{\left( a+\sqrt{x} \right)}^{3}}}\left[ \dfrac{1}{2}{{\left( x \right)}^{\dfrac{-1}{2}}} \right] \right] \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{2}\left[ \dfrac{1}{{{\left( a+\sqrt{x} \right)}^{2}}}\left( \dfrac{-1}{2} \right)\dfrac{1}{{{\left( x \right)}^{\dfrac{3}{2}}}}-\dfrac{2}{\sqrt{x}}.\dfrac{1}{{{\left( a+\sqrt{x} \right)}^{3}}}\left[ \dfrac{1}{2\sqrt{x}} \right] \right] \\
\end{align}\]
Cancelling the like terms, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{2}\left[ \dfrac{1}{{{\left( a+\sqrt{x} \right)}^{2}}}\left( \dfrac{-1}{2} \right)\dfrac{1}{{{\left( x \right)}^{\dfrac{3}{2}}}}-\dfrac{1}{x{{\left( a+\sqrt{x} \right)}^{3}}} \right]\]
Opening the bracket, we get
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{-1}{2} \right).\dfrac{1}{{{\left( a+\sqrt{x} \right)}^{2}}}\left( \dfrac{-1}{2} \right)\dfrac{1}{{{\left( x \right)}^{\dfrac{3}{2}}}}-\left( \dfrac{-1}{2} \right).\dfrac{1}{x{{\left( a+\sqrt{x} \right)}^{3}}} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{4{{\left( a+\sqrt{x} \right)}^{2}}{{\left( x \right)}^{\dfrac{3}{2}}}}+\dfrac{1}{2x{{\left( a+\sqrt{x} \right)}^{3}}} \\
\end{align}\]
Taking the LCM, we get
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( a+\sqrt{x} \right)+2\sqrt{x}}{4{{\left( a+\sqrt{x} \right)}^{3}}{{\left( x \right)}^{\dfrac{3}{2}}}} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( a+3\sqrt{x} \right)}{4{{\left( x \right)}^{\dfrac{3}{2}}}}\times \dfrac{1}{{{\left( a+\sqrt{x} \right)}^{3}}} \\
\end{align}\]
From the given expression we have \[y=\dfrac{1}{a+\sqrt{x}}\] , substituting this value in above equation, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( a+3\sqrt{x} \right)}{4{{\left( x \right)}^{\dfrac{3}{2}}}}\times {{y}^{3}}\]
This is the required answer.

Note: The way to solve the first and second order derivative is using the formula, $\dfrac{d}{dx}\left( \dfrac{1}{u(x)} \right)=-\dfrac{u'(x)}{u{{(x)}^{2}}}$ . In this method also we will get the same result.