What is the derivative of y=(x/(3.3)+(3.3)/x)(x^2+1) ?

The derivative of y=(x/(3.3)+(3.3)/x)(x^2+1) is 0.90909…x^2-(3.3)/(x^2)+3.60303…

What is the first derivative of y=(x/(3.3)+(3.3)/x)(x^2+1) ?

The first derivative of y=(x/(3.3)+(3.3)/x)(x^2+1) is 0.90909…x^2-(3.3)/(x^2)+3.60303…

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derivative of

y = (\frac{x}{3.3} + \frac{3.3}{x}) (x^{2} + 1)

0.90909 \dots x^{2} - \frac{3.3}{x ^{2}} + 3.60303 \dots

Solution steps

\frac{d}{d x} ((\frac{x}{3.3} + \frac{3.3}{x}) (x^{2} + 1))

Simplify

(\frac{x}{3.3} + \frac{3.3}{x}) (x^{2} + 1) : (0.30303 \dots x + \frac{3.3}{x}) (x^{2} + 1)

= \frac{d}{d x} ((0.30303 \dots x + \frac{3.3}{x}) (x^{2} + 1))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 0.30303 \dots x + \frac{3.3}{x}, g = x^{2} + 1

= \frac{d}{d x} (0.30303 \dots x + \frac{3.3}{x}) (x^{2} + 1) + \frac{d}{d x} (x^{2} + 1) (0.30303 \dots x + \frac{3.3}{x})

\frac{d}{d x} (0.30303 \dots x + \frac{3.3}{x}) = 0.30303 \dots - \frac{3.3}{x ^{2}}

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

= (0.30303 \dots - \frac{3.3}{x ^{2}}) (x^{2} + 1) + 2 x (0.30303 \dots x + \frac{3.3}{x})

Simplify

(0.30303 \dots - \frac{3.3}{x ^{2}}) (x^{2} + 1) + 2 x (0.30303 \dots x + \frac{3.3}{x}) : 0.90909 \dots x^{2} - \frac{3.3}{x ^{2}} + 3.60303 \dots

= 0.90909 \dots x^{2} - \frac{3.3}{x ^{2}} + 3.60303 \dots

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What is the derivative of y=(x/(3.3)+(3.3)/x)(x^2+1) ?
The derivative of y=(x/(3.3)+(3.3)/x)(x^2+1) is 0.90909…x^2-(3.3)/(x^2)+3.60303…
What is the first derivative of y=(x/(3.3)+(3.3)/x)(x^2+1) ?
The first derivative of y=(x/(3.3)+(3.3)/x)(x^2+1) is 0.90909…x^2-(3.3)/(x^2)+3.60303…