Question

differentiate y= (x²-3x+3)(x²+2x-1)

Answer 1

On expanding,
y= x⁴ + 2x³ - x² - 3x³ - 6x² + 3x + 3x² + 6x - 3
y= x⁴ - x³ - 4x² - 9x - 3
dy/dx = 4x³ - 3x² - 8x + 9
Hope this helps you.

Answer 2

The differentiation of y is $4x^{3} -3x^{2} -8x+9$.

Step-by-step explanation:

According to the given information is,

the function is given as,

y= (x²+2x-1)

Now, the following function y is in the multiplication format that is, when z = ab, where a and b are functions of x, we get,

$\frac{d}{dx} (z) = a\frac{d}{dx}(b) + b\frac{d}{dx} (a)$

Applying this rule on the given function y, we get,

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

This gives,

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

This happens because $\frac{d}{dx } (x^{n}) = nx^{n-1}$ , that is, when n is a non-zero power of x, the differentiation of x raised to that power n would be n multiplied by x to the power one subtracted from n.

Then, we get,

$(x^{2} -3x+3)(2x+2) + (x^{2} +2x-1)(2x-3)\\=2x^{3} +2x^{2} -6x^{2} -6x+6x+6+2x^{3} -3x^{2} +4x^{2} -6x-2x+3\\=4x^{3} -3x^{2} -8x+9$

Thus, the differentiation of y is $4x^{3} -3x^{2} -8x+9$.

Learn more here

https://brainly.in/question/8960809

Learn more

https://brainly.in/question/8628519