What is the derivative of x(x^4+9^3) ?

The derivative of x(x^4+9^3) is (x^4+9)^3+12x^4(x^4+9)^2

What is the first derivative of x(x^4+9^3) ?

The first derivative of x(x^4+9^3) is (x^4+9)^3+12x^4(x^4+9)^2

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\frac{d}{d x} (x (x^{4} + 9)^{3})

(x^{4} + 9)^{3} + 12 x^{4} (x^{4} + 9)^{2}

Solution steps

\frac{d}{d x} (x (x^{4} + 9)^{3})

Apply the Product Rule:

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

f = x, g = (x^{4} + 9)^{3}

= \frac{d x}{d x} (x^{4} + 9)^{3} + \frac{d}{d x} ((x^{4} + 9)^{3}) x

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

\frac{d}{d x} ((x^{4} + 9)^{3}) = 12 x^{3} (x^{4} + 9)^{2}

= 1 \cdot (x^{4} + 9)^{3} + 12 x^{3} (x^{4} + 9)^{2} x

Simplify

1 \cdot (x^{4} + 9)^{3} + 12 x^{3} (x^{4} + 9)^{2} x : (x^{4} + 9)^{3} + 12 x^{4} (x^{4} + 9)^{2}

= (x^{4} + 9)^{3} + 12 x^{4} (x^{4} + 9)^{2}

\int ln^{2} (\frac{1}{2} x) d x

\frac{( 12 s + 8 )}{( s - 1 ) ^{5}}

(1 - x) y^{^{'}} = y^{2}

\frac{- 8}{5 x + 5}

\frac{d}{d x} (tan (2))

What is the derivative of x(x^4+9^3) ?
The derivative of x(x^4+9^3) is (x^4+9)^3+12x^4(x^4+9)^2
What is the first derivative of x(x^4+9^3) ?
The first derivative of x(x^4+9^3) is (x^4+9)^3+12x^4(x^4+9)^2