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Tarun
2

g(x) is the inverse of f(x)

y=f(x); g(y)=x; f(x)=√(x^{2} + 6x); x>=0

f'(x) = (1/2).(2.x + 6)/√(x^{2} + 6x)

b=4 => y=4

√(x^{2} + 6x) = 4

x^{2}+6x-16 = 0

x = (-6 ± √(36+4(16)))/2 => x = 2

=> g(4) = 2; (x,y)=(2,4)

d(g(y))/dx = d(x)/dx => g'(y).y'=1 => g'(f(x)) = 1/f'(x)

=> g'(4) = 1/f'(2) = 1/(5/4) = 4/5

Mahesh Godavarti
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Very elegantly done.
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Sangeetha Pulapaka
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f(x) = √(x^{2}+ 6x)

Replace f(x) with y.

y = √(x^{2} + 6x)

Swap positions of x and y.

x = √(y^{2}+ 6y)

Square both sides.

x^{2} = y^{2}+6y

x^{2}+9 = y^{2}+6y+9

Factorize right side.

x^{2}+9 = (y +3)^{2}

Take square root of both sides.

√(x^{2}+ 9) = y + 3

Subtract 3 from both sides.

y = -3 + √(x^{2}+ 9)

Replace y with g(x).

g(x) = -3 + √(x^{2} + 9)

Plug in value: x = 4

Simplify.

g(4) = -3 + √(25)

Simplify.

g(4) = -3 + 5

Simplify.

g(4) = 2

-----

Use properties of roots: √(a+b) = (a+b)^{1/2}

g(x) = -3 + (x^{2} + 9)^{1/2}

Take derivative of both sides.

[d/dx] [g(x)] = [d/dx] [-3 + (x^{2} + 9)^{1/2}]

Distribute [d/dx] into bracket (right side).

[d/dx] [g(x)] = [d/dx] [3] + (1/2) (x^{2}+ 9)^{-1/2} [d/dx] [x^{2} + 9]

Evaluate remaining derivatives.

g'(x) = 0 + (1/2) (x^{2} + 9)^{-1/2} (2x)

Simplify.

g'(x) = (x) (x^{2} + 9)^{-1/2}

Use properties of fractional exponents: (a+b)^{-1/2} = [1 / √(a+b)]

Plug in value: b = 4.

g'(4) = (4) / [√(4^{2} + 9)]

Simplify.

g'(4) = (4) / [√(25)]

Simplify.

g'(4) = 4/5

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