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Question: Find the derivative of \({(3{x^2} - 7x + 3)^{\dfrac{5}{2}}}\) with respect to x....

Find the derivative of (3x27x+3)52{(3{x^2} - 7x + 3)^{\dfrac{5}{2}}} with respect to x.

Explanation

Solution

This type of question can be solved by using basic differentiation formulas. Here for a given function (3x27x+3)52{(3{x^2} - 7x + 3)^{\dfrac{5}{2}}}, use the chain rule of differential calculus for finding its derivative. Chain rule of differential calculus is given by dydx=dydududx\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}. Let’s assume u=3x27x+3u = 3{x^2} - 7x + 3. Use the power rule formula ddx(xn)=nxn1\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}} for finding derivation. Simplify it to get the final derivative of the given function (3x27x+3)52{(3{x^2} - 7x + 3)^{\dfrac{5}{2}}}.

Complete step-by-step answer:
Here the given function is (3x27x+3)52{(3{x^2} - 7x + 3)^{\dfrac{5}{2}}}.
Let’s say function y=(3x27x+3)52y = {(3{x^2} - 7x + 3)^{\dfrac{5}{2}}}.
Derivatives of such function can be obtained by using the chain rule of differential calculus.
Chain rule of differential calculus is given by dydx=dydududx\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}.
Now for given function y=(3x27x+3)52y = {(3{x^2} - 7x + 3)^{\dfrac{5}{2}}}, let’s assume u=3x27x+3u = 3{x^2} - 7x + 3.
So, y=(u)52y = {(u)^{\dfrac{5}{2}}}
Taking derivation of function y with respect x on both side of equation,
dydx=ddx(u52)\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}({u^{\dfrac{5}{2}}})
Now using chain rule of differential calculus,
dydx=ddx(u52)=ddu(u52)dudx\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}({u^{\dfrac{5}{2}}}) = \dfrac{d}{{du}}({u^{\dfrac{5}{2}}}) \cdot \dfrac{{du}}{{dx}}
Using basic derivation formula, ddx(xn)=nxn1\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}
So, ddu(u52)=52u(521)\dfrac{d}{{du}}({u^{\dfrac{5}{2}}}) = \dfrac{5}{2} \cdot {u^{(\dfrac{5}{2} - 1)}}
So, ddu(u52)=52u32\dfrac{d}{{du}}({u^{\dfrac{5}{2}}}) = \dfrac{5}{2} \cdot {u^{\dfrac{3}{2}}}
Put equation of u, as u=3x27x+3u = 3{x^2} - 7x + 3,
So, ddu(u52)=52(3x27x+3)32\dfrac{d}{{du}}({u^{\dfrac{5}{2}}}) = \dfrac{5}{2} \cdot {(3{x^2} - 7x + 3)^{\dfrac{3}{2}}}
And for dudx\dfrac{{du}}{{dx}},
dudx=ddx(3x27x+3)\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}(3{x^2} - 7x + 3)
Using sum of derivatives rule, ddx(f(x)+g(x))=ddx(f(x))+ddx(g(x))\dfrac{d}{{dx}}(f(x) + g(x)) = \dfrac{d}{{dx}}(f(x)) + \dfrac{d}{{dx}}(g(x))
So, dudx=ddx(3x2)+ddx(7x)+ddx(3)\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}(3{x^2}) + \dfrac{d}{{dx}}( - 7x) + \dfrac{d}{{dx}}(3)
dudx=ddx(3x2)+ddx(7x)+ddx(3)\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}(3{x^2}) + \dfrac{d}{{dx}}( - 7x) + \dfrac{d}{{dx}}(3)
As 3 and -7 are constant terms with respect to x, so taking the constant term out of derivation, as ddx(af(x))=addx(f(x))\dfrac{d}{{dx}}(a \cdot f(x)) = a \cdot \dfrac{d}{{dx}}(f(x)),
So, dudx=3ddx(x2)7ddx(x)+3ddx(1)\dfrac{{du}}{{dx}} = 3 \cdot \dfrac{d}{{dx}}({x^2}) - 7 \cdot \dfrac{d}{{dx}}(x) + 3 \cdot \dfrac{d}{{dx}}(1)
As we know that derivation of any constant term is zero, means ddx(a)=0\dfrac{d}{{dx}}(a) = 0
And using the basic derivation formula, ddx(xn)=nxn1\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}
So, dudx=32x217x11+3(0)\dfrac{{du}}{{dx}} = 3 \cdot 2{x^{2 - 1}} - 7 \cdot {x^{1 - 1}} + 3 \cdot (0)
Simplifying the above terms,
dudx=6x7x0+0\dfrac{{du}}{{dx}} = 6x - 7 \cdot {x^0} + 0
So, dudx=6x7(1)\dfrac{{du}}{{dx}} = 6x - 7 \cdot (1)
So, dudx=6x7\dfrac{{du}}{{dx}} = 6x - 7
Now putting the value of ddu(u52)\dfrac{d}{{du}}({u^{\dfrac{5}{2}}}) and dudx\dfrac{{du}}{{dx}} in dydx=ddu(u52)dudx\dfrac{{dy}}{{dx}} = \dfrac{d}{{du}}({u^{\dfrac{5}{2}}}) \cdot \dfrac{{du}}{{dx}}equation,
So, dydx=52(3x27x+3)32(6x7)\dfrac{{dy}}{{dx}} = \dfrac{5}{2} \cdot {(3{x^2} - 7x + 3)^{\dfrac{3}{2}}} \cdot (6x - 7)
Arranging the terms, dydx=52(6x7)(3x27x+3)32\dfrac{{dy}}{{dx}} = \dfrac{5}{2} \cdot (6x - 7) \cdot {(3{x^2} - 7x + 3)^{\dfrac{3}{2}}}.

So the derivative of the function (3x27x+3)52{(3{x^2} - 7x + 3)^{\dfrac{5}{2}}} with respect to x is given by dydx=52(6x7)(3x27x+3)32\dfrac{{dy}}{{dx}} = \dfrac{5}{2} \cdot (6x - 7) \cdot {(3{x^2} - 7x + 3)^{\dfrac{3}{2}}}.

Note: If the given function is (3x27x+3)x{(3{x^2} - 7x + 3)^x}, then it is neither a power function form (xn)({x^n}) nor a exponential functional form (ax)({a^x}). So formulas for differentiation of these forms cannot be used. To derivative of such function y=(3x27x+3)xy = {(3{x^2} - 7x + 3)^x}, take natural logarithm on both sides of the equation.
So, ln(y)=ln((3x27x+3)x)\ln (y) = \ln ({(3{x^2} - 7x + 3)^x}).
Use properties of logarithmic functions to expand term of right side,
Taking derivative of above equation on both side with respect to x,
Use chain rule of derivation, dydx=dydududx\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \cdot \dfrac{{du}}{{dx}}, product rule of derivation ddx(f(x)g(x))=(ddxf(x))g(x)+f(x)(ddxg(x))\dfrac{d}{{dx}}(f(x) \cdot g(x)) = (\dfrac{d}{{dx}}f(x)) \cdot g(x) + f(x) \cdot (\dfrac{d}{{dx}}g(x)) and some basic formula of derivation as used in above solution and simplify the solution to find derivative of function (3x27x+3)x{(3{x^2} - 7x + 3)^x}.