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

The derivative of (9x^3+3x^2^4) is 4(9x^3+3x^2)^3(27x^2+6x)

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

The first derivative of (9x^3+3x^2^4) is 4(9x^3+3x^2)^3(27x^2+6x)

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

4 (9 x^{3} + 3 x^{2})^{3} (27 x^{2} + 6 x)

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (9 x^{3} + 3 x^{2}) = 27 x^{2} + 6 x

= 4 (9 x^{3} + 3 x^{2})^{3} (27 x^{2} + 6 x)

- y^{^{'}} - 5 y = - t

\frac{d y}{d x} + 10 (tan (10 x)) y = cos (10 x)

\frac{d y}{d x} = 4 x^{3} y

\int \frac{D ( 5 x ^{2} + 5 y ^{2} ) - 1}{2} d A

x y^{^{'}} + 2 y = 3 sin (x)

What is the derivative of (9x^3+3x^2^4) ?
The derivative of (9x^3+3x^2^4) is 4(9x^3+3x^2)^3(27x^2+6x)
What is the first derivative of (9x^3+3x^2^4) ?
The first derivative of (9x^3+3x^2^4) is 4(9x^3+3x^2)^3(27x^2+6x)