Question

x(1 + x)³ find the derivative​

Answer 1

Answer:

4x³ + 9x² + 6x + 1

Step-by-step explanation:

To find ---> Derivative of x ( 1 + x )³

Solution---> Let y = x ( 1 + x )³

We have an identity

(a + b )³ = a³ + b³ + 3ab( a+ b ) , we get

=> y = x { 1 + x³ + 3x ( 1 + x ) }

= x ( 1 + x³ + 3x + 3x² )

= x + x x³ + 3 x x + 3 x x²

We have a law of exponent

aᵐ aⁿ = aᵐ⁺ⁿ , applying it , we get

= x + x¹⁺³ + 3x¹⁺¹ + 3 x¹⁺²

= x + x⁴ + 3x² + 3x³

=> y = x⁴ + 3x³ + 3x² + x

Differentiating with respect to x , we get

=> dy/dx = d/dx ( x⁴ + 3x³ + 3x² + x )

We have a formula of differentiation

d / dx (xⁿ) = nxⁿ⁻¹ , applying it , we get

= d/dx (x⁴) + d/dx (3x³) +d/dx (3x²) + d/dx (x)

= 4x⁴⁻¹ + 3 (3)x³⁻¹ + 3 (2x²⁻¹) + 1

dy /dx= 4x³ + 9x² + 6x + 1

Answer 2

Question :----- Find the derivative of x(1+x)³

Formula to be used :------

• (a+b)³ = a³+b³+3ab(a+b)
• d/dx(x^n+AX+C) = nx+A
• d/dx(x^n) = n×x^(n-1)
• x^0 = 1

Solving the Equation first ,

x(1+x)³

→ x[1+x³+3x(1+x)]

→ x[x³+3x²+3x+1]

→ [x⁴+3x³+3x²+x]

Now , differentiate the equation we get,

$\frac{d}{dx} ( {x}^{4} + 3{x}^{3} + 3{x}^{2} + x) \\ \\ 4x^{4 - 1} + 3 \times 3 {x}^{3 - 1} + 3 \times 2 {x}^{2 - 1} + 1 {x}^{1 - 1} \\ \\ 4 {x}^{3} + 9 {x}^{2} + 6x + 1$

(Hope it helps you)