What is the derivative of (x^2-3x+1^4) ?

The derivative of (x^2-3x+1^4) is 4(x^2-3x+1)^3(2x-3)

What is the first derivative of (x^2-3x+1^4) ?

The first derivative of (x^2-3x+1^4) is 4(x^2-3x+1)^3(2x-3)

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\frac{d}{d x} ((x^{2} - 3 x + 1)^{4})

4 (x^{2} - 3 x + 1)^{3} (2 x - 3)

Solution steps

\frac{d}{d x} ((x^{2} - 3 x + 1)^{4})

Apply the chain rule:

4 (x^{2} - 3 x + 1)^{3} \frac{d}{d x} (x^{2} - 3 x + 1)

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

\frac{d}{d x} (x^{2} - 3 x + 1) = 2 x - 3

= 4 (x^{2} - 3 x + 1)^{3} (2 x - 3)

5 x, \frac{5}{3} x, 14 - x^{2}

9 e^{x}

y (X) = \frac{1}{2 cos ^{2} ( X )}

y = 4 x ln (x) + (3^{x}) - (x^{3})

\int \frac{1}{( 256 + x ^{2} ) ^{\frac{3}{2}}} d x

What is the derivative of (x^2-3x+1^4) ?
The derivative of (x^2-3x+1^4) is 4(x^2-3x+1)^3(2x-3)
What is the first derivative of (x^2-3x+1^4) ?
The first derivative of (x^2-3x+1^4) is 4(x^2-3x+1)^3(2x-3)