What is the derivative of 3x(18x^4+(13)/(x+1)) ?

The derivative of 3x(18x^4+(13)/(x+1)) is (3(90x^6+180x^5+90x^4+13))/((x+1)^2)

What is the first derivative of 3x(18x^4+(13)/(x+1)) ?

The first derivative of 3x(18x^4+(13)/(x+1)) is (3(90x^6+180x^5+90x^4+13))/((x+1)^2)

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derivative of

3 x (18 x^{4} + \frac{13}{x + 1})

\frac{3 ( 90 x ^{6} + 180 x ^{5} + 90 x ^{4} + 13 )}{( x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (3 x (18 x^{4} + \frac{13}{x + 1}))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d x} (x (18 x^{4} + \frac{13}{x + 1}))

Apply the Product Rule:

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

f = x, g = 18 x^{4} + \frac{13}{x + 1}

= 3 (\frac{d x}{d x} (18 x^{4} + \frac{13}{x + 1}) + \frac{d}{d x} (18 x^{4} + \frac{13}{x + 1}) x)

\frac{d x}{d x} = 1

\frac{d}{d x} (18 x^{4} + \frac{13}{x + 1}) = 72 x^{3} - \frac{13}{( x + 1 ) ^{2}}

= 3 (1 \cdot (18 x^{4} + \frac{13}{x + 1}) + (72 x^{3} - \frac{13}{( x + 1 ) ^{2}}) x)

Simplify

3 (1 \cdot (18 x^{4} + \frac{13}{x + 1}) + (72 x^{3} - \frac{13}{( x + 1 ) ^{2}}) x) : \frac{3 ( 90 x ^{6} + 180 x ^{5} + 90 x ^{4} + 13 )}{( x + 1 ) ^{2}}

= \frac{3 ( 90 x ^{6} + 180 x ^{5} + 90 x ^{4} + 13 )}{( x + 1 ) ^{2}}

\int 1 + cos (2 x - 2) d x

- 8 sin (t)

(12 y^{2} - x y) d x + x^{2} d y = 0

f (x) = \frac{3}{x}, (8, \frac{3}{8})

\int \frac{4}{5 - 3 x} d x

What is the derivative of 3x(18x^4+(13)/(x+1)) ?
The derivative of 3x(18x^4+(13)/(x+1)) is (3(90x^6+180x^5+90x^4+13))/((x+1)^2)
What is the first derivative of 3x(18x^4+(13)/(x+1)) ?
The first derivative of 3x(18x^4+(13)/(x+1)) is (3(90x^6+180x^5+90x^4+13))/((x+1)^2)