calculus question (1 Viewer)

blacktown_ · Jan 4, 2013

find the exact value of f''(2) if f(x)= x (3x-4)^1/2

Focus is Key · Jan 5, 2013

Hint: f'(x)= 1(3x-4)^1/2+x.1/2(3x-4)^-1/2.3

Where I put the dot is times. Then simplfy, differentiate again for f''(x), then subsitute 2 into the second derivative.

Sy123 · Jan 5, 2013

blacktown_ said:
find the exact value of f''(2) if f(x)= x (3x-4)^1/2

Always look for shortcuts where ever possible:

$f(x)=x\sqrt{3x-4} = \sqrt{x^2(3x-4)} = \sqrt{3x^3-4x^2}$

This makes differentiating easier:

$f'(x)= \frac{1}{2}(9x^2-8x) (3x^3-4x^2)^{\frac{-1}{2}}$

$f'(x)=\frac{9x^2-8x}{2\sqrt{4x^3-4x^2}}$

unfortunately we can't take the same shortcut here:

$f''(x) = \frac{(18x-8) \cdot 2\sqrt{3x^3-4x^2} - 2(9x^2-8x)(3x^3-4x^2)^{-1/2}}{4(4x^3-4x^2)}$

Sub in x=2

$f''(2)= \dots$

Whatever it might be

calculus question (1 Viewer)

blacktown_

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Focus is Key

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Sy123

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