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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

\int \frac{e ^{2 y}}{e ^{y} - 1} d y

\frac{d y}{d x} - x y^{2} = 2 x y

\int t cos^{5} (t^{2}) d t

y = 1 + 2 e^{8 x}

n = 1 \sum \infty \frac{1}{7 n ^{2} + n + 6}

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