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Question: Find \[\dfrac{dy}{dx}\] of \[y={{\left( \dfrac{x-5}{2x+1} \right)}^{3}}\]...

Find dydx\dfrac{dy}{dx} of y=(x−52x+1)3y={{\left( \dfrac{x-5}{2x+1} \right)}^{3}}

Explanation

Solution

Hint: To solve the above problem we have to know the basic derivatives of x3{{x}^{3}} and internal derivatives. After writing the derivatives rewrite the equation with the derivatives of the function. ddx(y)n=n.yn−1dydx\dfrac{d}{dx}{{\left( y \right)}^{n}}=n.{{y}^{n-1}}\dfrac{dy}{dx} We can see one function is inside another we have to find internal derivative.

Complete step-by-step answer:
y=(x−52x+1)3y={{\left( \dfrac{x-5}{2x+1} \right)}^{3}}. . . . . . . . . . . . . . . . . . . . . (a)
ddx(y)n=n.yn−1dydx\dfrac{d}{dx}{{\left( y \right)}^{n}}=n.{{y}^{n-1}}\dfrac{dy}{dx}. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
ddx(x)3=3x2\dfrac{d}{dx}{{\left( x \right)}^{3}}=3{{x}^{2}}. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Substituting (1) and (2) in (a) we get,
Therefore derivative of the given function is,
dydx=3(x−52x+1)2(ddx(x−52x+1))\dfrac{dy}{dx}=3{{\left( \dfrac{x-5}{2x+1} \right)}^{2}}\left( \dfrac{d}{dx}\left( \dfrac{x-5}{2x+1} \right) \right)
Now the quotient rule is applied, By writing the derivatives we get,
Further solving we get the derivative of the function as
dydx=3(x−52x+1)2(2x+1(ddx(x−5)−(x−5)(ddx(2x+1))(2x+1)2)\dfrac{dy}{dx}=3{{\left( \dfrac{x-5}{2x+1} \right)}^{2}}\left( \dfrac{2x+1(\dfrac{d}{dx}\left( x-5 \right)-(x-5)(\dfrac{d}{dx}\left( 2x+1 \right))}{{{(2x+1)}^{2}}} \right)
By solving we get,
dydx=3(x−52x+1)2((2x+1)(1)−(x−5)(2)(2x+1)2)\dfrac{dy}{dx}=3{{\left( \dfrac{x-5}{2x+1} \right)}^{2}}\left( \dfrac{(2x+1)(1)-(x-5)(2)}{{{(2x+1)}^{2}}} \right)
dydx=3(x−52x+1)2((2x+1)−(2x−10)(2x+1)2)\dfrac{dy}{dx}=3{{\left( \dfrac{x-5}{2x+1} \right)}^{2}}\left( \dfrac{(2x+1)-(2x-10)}{{{(2x+1)}^{2}}} \right)
dydx=3(x−52x+1)2(11(2x+1)2)\dfrac{dy}{dx}=3{{\left( \dfrac{x-5}{2x+1} \right)}^{2}}\left( \dfrac{11}{{{(2x+1)}^{2}}} \right)

Note: In the above problem we have solved the derivative of a cubic expression and we found out the internal derivative by quotient rule. In this case we have to find an internal derivative. Further solving for dydx\dfrac{dy}{dx}made us towards a solution. If we are doing derivative means we are finding the slope of a function. Care should be taken while doing calculations.