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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

\frac{\partial}{\partial x} (x \cdot y^{2} \cdot e^{z})

y = cos^{3} (π t - 16)

f (x) = x^{4} + 2 x^{2} - x, (1, 2)

\frac{d}{d x} (x, x^{2} - y, x + y^{2})

\int \frac{5}{( 8 z - 1 ) ^{\frac{7}{8}}} d z

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