# If $a{y^4} = {(x + b)^5}$ then show that $5y{y''} = {(y')^2}$

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**Hint:**First we have to define the terms we need to solve in the problem. Since this problem is based on the method of the differentiation which is if we differentiate ${x^n}$ then we get $n{x^{n - 1}}$ so using this concept we will solve it further. Also, differentiation is with respect to the variables or values or numbers. This may also be called the inverse of the integration.

**Complete step-by-step solution:**

From the given question we have $a{y^4} = {(x + b)^5}$ first place the value $a$into the right hand side; then we get ${y^4} = \dfrac{{{{(x + b)}^5}}}{a}$ (a and b are constants here so after acting the differentiation does not affects the constant ) In next step turn the square values into square root to the right hand side or take common square root on both side we get; $y = {(\dfrac{1}{a})^{\dfrac{1}{4}}}{(x + b)^{\dfrac{5}{4}}}$ now we may assume that constant term into C

And hence $y = c{(x + b)^{\dfrac{5}{4}}}$ and since (${(\dfrac{1}{a})^{\dfrac{1}{4}}}$=C)

From the given definition we can further solve different ion of the value with respect to x;

First differentiation with respect to x we have; $y' = c \times \dfrac{5}{4}{(x + b)^{\dfrac{1}{4}}}$(the power will come down also the new power minus one) now again differentiation with respect to x we have;

${y''} = \dfrac{5}{4}c \times \dfrac{1}{4}{(x + b)^{\dfrac{{ - 3}}{4}}} $

$= \dfrac{{5c}}{{16}}{(x + b)^{\dfrac{{ - 3}}{4}}}$ (the power will come down also the new power minus one)

Hence, we have first and second differentiation of the given problem

Now from the problem take the left-hand side which is $5y{y''}$ and substitute the first and second differentiation values we get

$5y{y''} = 5c{(x + b)^{\tfrac{5}{4}}} \times \dfrac{{5c}}{{16}}{(x + b)^{\dfrac{{ - 3}}{4}}}$ and further solving by cancelling the terms we get;

$5yy{}'' = \dfrac{{25{c^2}}}{{16}}{(x + b)^{\dfrac{1}{2}}}$

Similarly, from the right side given equation ${(y')^2}$ and substitute the first differentiation we get;

${(y')^2} = {[c \times \dfrac{5}{4}{(x + b)^{\dfrac{1}{4}}}]^2} = \dfrac{{25{c^2}}}{{16}}{(x + b)^{\dfrac{1}{2}}}$

**Thus clear the right and left side equations given the same value and hence $5y{y''} = {(y')^2}$.**

**Note:**Here $x,y$are the variables so differentiation will only act on them and $a,b$are the constants so it won’t affect these values.

Also, differentiation of ${x^2} = 2{x^{2 - 1}} = 2x$ and ${x^{ - 2}} = - 2{x^{ - 3}}$;

Integration is the inverse function of differentiation. There is no other easy method to solve this problem.

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