What is the derivative of f(x)= 4/(3x^2)+1/(x^{2.1)} ?

The derivative of f(x)= 4/(3x^2)+1/(x^{2.1)} is -(2.66666…)/(x^3)-(2.1)/(x^{3.1)}

What is the first derivative of f(x)= 4/(3x^2)+1/(x^{2.1)} ?

The first derivative of f(x)= 4/(3x^2)+1/(x^{2.1)} is -(2.66666…)/(x^3)-(2.1)/(x^{3.1)}

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derivative of

f (x) = \frac{4}{3 x ^{2}} + \frac{1}{x ^{2.1}}

- \frac{2.66666 \dots}{x ^{3}} - \frac{2.1}{x ^{3.1}}

Solution steps

\frac{d}{d x} (\frac{4}{3 x ^{2}} + \frac{1}{x ^{2.1}})

Simplify

\frac{4}{3 x ^{2}} + \frac{1}{x ^{2.1}} : \frac{1.33333 \dots}{x ^{2}} + \frac{1}{x ^{2.1}}

= \frac{d}{d x} (\frac{1.33333 \dots}{x ^{2}} + \frac{1}{x ^{2.1}})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{1.33333 \dots}{x ^{2}}) + \frac{d}{d x} (\frac{1}{x ^{2.1}})

\frac{d}{d x} (\frac{1.33333 \dots}{x ^{2}}) = - \frac{2.66666 \dots}{x ^{3}}

\frac{d}{d x} (\frac{1}{x ^{2.1}}) = - \frac{2.1}{x ^{3.1}}

= - \frac{2.66666 \dots}{x ^{3}} - \frac{2.1}{x ^{3.1}}

\int \frac{( 8 + x + x ^{2} )}{x} d x

x \to 4 - lim (\frac{6 x}{x - 4})

f (x) = 3 sin^{2} (x)

2 y^{^{''}} + y^{^{'}} - 2 y = 0

x \to 5 lim (cos (x))

What is the derivative of f(x)= 4/(3x^2)+1/(x^{2.1)} ?
The derivative of f(x)= 4/(3x^2)+1/(x^{2.1)} is -(2.66666…)/(x^3)-(2.1)/(x^{3.1)}
What is the first derivative of f(x)= 4/(3x^2)+1/(x^{2.1)} ?
The first derivative of f(x)= 4/(3x^2)+1/(x^{2.1)} is -(2.66666…)/(x^3)-(2.1)/(x^{3.1)}