Question & Answer
QUESTION

Find the derivative of the following function. \[y=\sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}} + \dfrac{\sqrt[3]{{{x}^{-8}}\sqrt{{{x}^{9}}}}}{\sqrt{4x\sqrt{{{x}^{5}}}}}\]

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Hint:- To solve this question first we have to separate the given function then we will apply the quotient rule, sum rule, product rule and chain rule of composition of two functions to find the derivative of the given function.

Complete step-by-step solution -
To find the derivative of the function \[y=\sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}}+\dfrac{\sqrt[3]{{{x}^{-8}}\sqrt{{{x}^{9}}}}}{\sqrt{4x\sqrt{{{x}^{5}}}}}\], we will differentiate it with respect to the variable x.
We can write this function as a sum of two functions \[y=f\left( x \right)+g\left( x \right)\] and use sum rule of differentiation which states that if \[y=f\left( x \right)+g\left( x \right)\] then, we have \[\dfrac{dy}{dx}=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)\].
Substituting \[f\left( x \right)=\sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}},g\left( x \right)=\dfrac{\sqrt[3]{{{x}^{-8}}\sqrt{{{x}^{9}}}}}{\sqrt{4x\sqrt{{{x}^{5}}}}}\] in the above equation, we get \[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}} \right)+\dfrac{d}{dx}\left( \dfrac{\sqrt[3]{{{x}^{-8}}\sqrt{{{x}^{9}}}}}{\sqrt{4x\sqrt{{{x}^{5}}}}} \right).....\left( 1 \right)\].
To find the value of \[\dfrac{d}{dx}\left( \sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}} \right)\], we can write \[f\left( x \right)=\sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}}=u(v(x))\] as a composition of two functions.
We will use chain rule of differentiation of composition of two functions to find the derivative which states that if y is a composition of two functions \[u(x)\] and \[v(x)\] such that \[y=u(v(x))\], then \[\dfrac{dy}{dx}=\dfrac{du(v(x))}{dv(x)}\times \dfrac{dv(x)}{dx}\].
Substituting \[u\left( x \right)=\sqrt{x},v\left( x \right)=\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}\] in the above equation, we have \[\dfrac{d}{dx}f\left( x \right)=\dfrac{du\left( v\left( x \right) \right)}{dv\left( x \right)}\times \dfrac{dv\left( x \right)}{dx}=\dfrac{d\left( \sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}} \right)}{d\left( \dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}} \right)}\times \dfrac{d\left( \dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}} \right)}{dx}.....\left( 2 \right)\].
To find the value of \[\dfrac{d\left( \sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}} \right)}{d\left( \dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}} \right)}\], let’s assume \[t=\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}\].
Thus, we have \[\dfrac{d\left( \sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}} \right)}{d\left( \dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}} \right)}=\dfrac{d\left( \sqrt{t} \right)}{dt}\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=\dfrac{1}{2},b=0\] in the above equation, we have \[\dfrac{d\left( \sqrt{t} \right)}{dt}=\dfrac{1}{2\sqrt{t}}\].
Thus, we have \[\dfrac{d\left( \sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}} \right)}{d\left( \dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}} \right)}=\dfrac{d\left( \sqrt{t} \right)}{dt}=\dfrac{1}{2\sqrt{t}}=\dfrac{1}{2\sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}}}.....\left( 3 \right)\].
To find the value of \[\dfrac{d\left( \dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}} \right)}{dx}\], we can write \[v\left( x \right)=\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}\] as a quotient of two functions of the form \[\dfrac{a\left( x \right)}{b\left( x \right)}\].
We will use quotient rule of differentiation which states that if \[y=\dfrac{a(x)}{b(x)}\] then, we have \[\dfrac{dy}{dx}=\dfrac{b(x)a'(x)-a(x)b'(x)}{{{b}^{2}}(x)}\].
Substituting \[a\left( x \right)=\sqrt[5]{{{x}^{7}}}={{x}^{\dfrac{7}{5}}},b\left( x \right)=6{{x}^{3}}\] in the above equation, we have \[\dfrac{d}{dx}v\left( x \right)=\dfrac{b(x)a'(x)-a(x)b'(x)}{{{b}^{2}}(x)}=\dfrac{6{{x}^{3}}\times \dfrac{d}{dx}\left( {{x}^{\dfrac{7}{5}}} \right)-{{x}^{\dfrac{7}{5}}}\times \dfrac{d}{dx}\left( 6{{x}^{3}} \right)}{{{\left( 6{{x}^{3}} \right)}^{2}}}.....\left( 4 \right)\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=6,n=3,b=0\] in the above equation, we have \[\dfrac{d}{dx}\left( 6{{x}^{3}} \right)=18{{x}^{2}}.....\left( 5 \right)\].
Substituting \[a=1,n=\dfrac{7}{5},b=0\] in the above equation, we have \[\dfrac{d}{dx}\left( {{x}^{\dfrac{7}{5}}} \right)=\dfrac{7}{5}{{x}^{\dfrac{2}{5}}}.....\left( 6 \right)\].
Substituting equation (5) and (6) in equation (4), we get \[\dfrac{d}{dx}v\left( x \right)=\dfrac{6{{x}^{3}}\times \dfrac{d}{dx}\left( {{x}^{\dfrac{7}{5}}} \right)-{{x}^{\dfrac{7}{5}}}\times \dfrac{d}{dx}\left( 6{{x}^{3}} \right)}{{{\left( 6{{x}^{3}} \right)}^{2}}}=\dfrac{6{{x}^{3}}\times \dfrac{7}{5}{{x}^{\dfrac{2}{5}}}-{{x}^{\dfrac{7}{5}}}\times 18{{x}^{2}}}{36{{x}^{6}}}\].
Solving the above equation by taking LCM and cutting out the common terms, we get \[\dfrac{d}{dx}v\left( x \right)=\dfrac{6{{x}^{3}}\times \dfrac{7}{5}{{x}^{\dfrac{2}{5}}}-{{x}^{\dfrac{7}{5}}}\times 18{{x}^{2}}}{36{{x}^{6}}}=\dfrac{6{{x}^{2}}\times {{x}^{\dfrac{2}{5}}}\left( \dfrac{7x}{5}-3x \right)}{36{{x}^{6}}}=\dfrac{{{x}^{\dfrac{2}{5}}}\times \left( \dfrac{-8x}{5} \right)}{6{{x}^{4}}}=\dfrac{-4{{x}^{\dfrac{2}{5}}}}{15{{x}^{3}}}.....\left( 7 \right)\].
Substituting equation (3) and (7) in equation (2), we get \[\dfrac{d}{dx}f\left( x \right)=\dfrac{d\left( \sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}} \right)}{d\left( \dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}} \right)}\times \dfrac{d\left( \dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}} \right)}{dx}=\dfrac{1}{2\sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}}}\times \dfrac{-4{{x}^{\dfrac{2}{5}}}}{15{{x}^{3}}}=\dfrac{-2{{x}^{\dfrac{2}{5}}}}{15{{x}^{3}}\sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}}}.....\left( 8 \right)\].
To find the value of \[\dfrac{d}{dx}\left( \dfrac{\sqrt[3]{{{x}^{-8}}\sqrt{{{x}^{9}}}}}{\sqrt{4x\sqrt{{{x}^{5}}}}} \right)\], we can write \[g\left( x \right)=\dfrac{\sqrt[3]{{{x}^{-8}}\sqrt{{{x}^{9}}}}}{\sqrt{4x\sqrt{{{x}^{5}}}}}\] as a quotient of two functions \[\dfrac{c\left( x \right)}{d\left( x \right)}\] where \[c\left( x \right)=\sqrt[3]{{{x}^{-8}}\sqrt{{{x}^{9}}}},d\left( x \right)=\sqrt{4x\sqrt{{{x}^{5}}}}\].
We can rewrite \[c\left( x \right)\] as \[c\left( x \right)=\sqrt[3]{{{x}^{-8}}\sqrt{{{x}^{9}}}}=\sqrt[3]{{{x}^{-8+\dfrac{9}{2}}}}=\sqrt[3]{{{x}^{\dfrac{-7}{2}}}}={{x}^{\dfrac{-7}{6}}}\].
We can rewrite \[d\left( x \right)\] as \[d\left( x \right)=\sqrt{4x\sqrt{{{x}^{5}}}}=2\sqrt{x\times {{x}^{\dfrac{5}{2}}}}=2\sqrt{{{x}^{1+\dfrac{5}{2}}}}=2\sqrt{{{x}^{\dfrac{7}{2}}}}=2{{x}^{\dfrac{7}{4}}}\].
We will use quotient rule of differentiation which states that if \[y=\dfrac{c(x)}{d(x)}\] then, we have \[\dfrac{dy}{dx}=\dfrac{d(x)c'(x)-c(x)d'(x)}{{{d}^{2}}(x)}\].
Substituting \[c\left( x \right)={{x}^{\dfrac{-7}{6}}},d\left( x \right)=2{{x}^{\dfrac{7}{4}}}\] in the above equation, we get \[\dfrac{d}{dx}g\left( x \right)=\dfrac{d(x)c'(x)-c(x)d'(x)}{{{d}^{2}}(x)}=\dfrac{2{{x}^{\dfrac{7}{4}}}\times \dfrac{d}{dx}\left( {{x}^{\dfrac{-7}{6}}} \right)-{{x}^{\dfrac{-7}{6}}}\times \dfrac{d}{dx}\left( 2{{x}^{\dfrac{7}{4}}} \right)}{{{\left( 2{{x}^{\dfrac{7}{4}}} \right)}^{2}}}.....\left( 9 \right)\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=\dfrac{-7}{6},b=0\] in the above equation, we get \[\dfrac{d}{dx}c\left( x \right)=\dfrac{d}{dx}\left( {{x}^{\dfrac{-7}{6}}} \right)=\dfrac{-7{{x}^{\dfrac{-13}{6}}}}{6}.....\left( 10 \right)\].
Substituting \[a=2,n=\dfrac{7}{4},b=0\] in the above equation, we get \[\dfrac{d}{dx}d\left( x \right)=\dfrac{d}{dx}\left( 2{{x}^{\dfrac{7}{4}}} \right)=2\times \dfrac{7}{4}{{x}^{\dfrac{3}{4}}}=\dfrac{7}{2}{{x}^{\dfrac{3}{4}}}.....\left( 11 \right)\].
Substituting equation (10) and (11) in equation (9), we get \[\dfrac{d}{dx}g\left( x \right)=\dfrac{2{{x}^{\dfrac{7}{4}}}\times \dfrac{d}{dx}\left( {{x}^{\dfrac{-7}{6}}} \right)-{{x}^{\dfrac{-7}{6}}}\times \dfrac{d}{dx}\left( 2{{x}^{\dfrac{7}{4}}} \right)}{{{\left( 2{{x}^{\dfrac{7}{4}}} \right)}^{2}}}=\dfrac{2{{x}^{\dfrac{7}{4}}}\times \dfrac{-7{{x}^{\dfrac{-13}{6}}}}{6}-{{x}^{\dfrac{-7}{6}}}\times \dfrac{7}{2}{{x}^{\dfrac{3}{4}}}}{4{{x}^{\dfrac{7}{2}}}}\].
Solving the above equation, we get \[\dfrac{d}{dx}g\left( x \right)=\dfrac{2{{x}^{\dfrac{7}{4}}}\times \dfrac{-7{{x}^{\dfrac{-13}{6}}}}{6}-{{x}^{\dfrac{-7}{6}}}\times \dfrac{7}{2}{{x}^{\dfrac{3}{4}}}}{4{{x}^{\dfrac{7}{2}}}}=\dfrac{\dfrac{-7}{3}{{x}^{\dfrac{7}{4}-\dfrac{13}{6}}}-\dfrac{7}{2}{{x}^{\dfrac{3}{4}-\dfrac{7}{6}}}}{4{{x}^{\dfrac{7}{2}}}}=\dfrac{\dfrac{-7}{3}{{x}^{\dfrac{-5}{12}}}-\dfrac{7}{2}{{x}^{-\dfrac{5}{12}}}}{4{{x}^{\dfrac{7}{2}}}}=\dfrac{\dfrac{-35}{6}{{x}^{-\dfrac{5}{12}}}}{4{{x}^{\dfrac{7}{2}}}}=\dfrac{-35}{24}{{x}^{\dfrac{-47}{12}}}.....\left( 12 \right)\].
Substituting equation (8) and (12) in equation (1), we get \[\dfrac{dy}{dx}=\dfrac{-2{{x}^{\dfrac{2}{5}}}}{15{{x}^{3}}\sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}}}+\dfrac{-35}{24}{{x}^{\dfrac{-47}{12}}}\].
Thus, differentiation of the function \[y=\sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}}+\dfrac{\sqrt[3]{{{x}^{-8}}\sqrt{{{x}^{9}}}}}{\sqrt{4x\sqrt{{{x}^{5}}}}}\] is \[\dfrac{dy}{dx}=\dfrac{-2{{x}^{\dfrac{2}{5}}}}{15{{x}^{3}}\sqrt{\dfrac{\sqrt[5]{{{x}^{7}}}}{6{{x}^{3}}}}}+\dfrac{-35}{24}{{x}^{\dfrac{-47}{12}}}\].

Note: The first derivative of any function signifies the slope of the function. It’s necessary to use the chain rule of composition of two functions to solve the above question. Otherwise, we won’t be able to find the derivative of a given function.