Question

 y=4x²-3x+1 y` -?  y= cosx+sinx y` -? ʄ(x) = x2  * ( x +3 )2

Answer 1

1) 8x-3

2) cosx-sinx

3) 2*x*(x-3)*(2x-3)

Answer 2

To find the derivative of the given functions, we can apply the power rule and the sum rule of differentiation. Let's calculate the derivatives:

1. y = 4x² - 3x + 1

To find y', the derivative of y with respect to x, we differentiate each term separately:

y' = d/dx (4x²) - d/dx (3x) + d/dx (1)

Applying the power rule, we have:

y' = 8x - 3

Therefore, the derivative of y = 4x² - 3x + 1 is y' = 8x - 3.

2. y = cos(x) + sin(x)

To find y', we differentiate each term using the sum rule:

y' = d/dx (cos(x)) + d/dx (sin(x))

The derivative of cosine is negative sine, and the derivative of sine is cosine:

y' = -sin(x) + cos(x)

So, the derivative of y = cos(x) + sin(x) is y' = -sin(x) + cos(x).

3. f(x) = x² * (x + 3)²

To find f'(x), we can apply the product rule and the chain rule:

f'(x) = (d/dx (x²)) * (x + 3)² + x² * (d/dx (x + 3)²)

Applying the power rule, we have:

f'(x) = (2x) * (x + 3)² + x² * (2(x + 3))

Simplifying further:

f'(x) = 2x(x + 3)² + 2x²(x + 3)

f'(x) = 2x(x + 3)[(x + 3) + x²]

f'(x) = 2x(x + 3)(x + 3 + x²)

Therefore, the derivative of ʄ(x) = x² * (x + 3)² is f'(x) = 2x(x + 3)(x + 3 + x²).