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3 years ago

F(x) = 4√(2x³-1) F'(x) =.....?

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

6 0

Answer:

$\large\boxed{f'(x)=\dfrac{12x^2}{\sqrt{2x^3-1}}}$

Step-by-step explanation:

$f(x)=4\sqrt{2x^3-1}=4\left(2x^3-1\right)^\frac{1}{2}\\\\f'(x)=4\cdot\dfrac{1}{2}(2x^3-1)^{-\frac{1}{2}}\cdot3\cdot2x^2=\dfrac{12x^2}{(2x^3-1)^\frac{1}{2}}=\dfrac{12x^2}{\sqrt{2x^3-1}}\\\\\text{used}\\\\\sqrt{a}=a^\frac{1}{2}\\\\\bigg[f\left(g(x)\right)\bigg]'=f'(g(x))\cdot g'(x)\\\\\bigg[nf(x)\bigg]'=nf'(x)\\\\(x^n)'=nx^{n-1}$

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Now assume this is true for $n=k$ , i.e.

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and under this hypothesis show that it's also true for $n=k+1$ . You have

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