Question

d/dx(x2+3x) whole square​

Answer 1

Question :

Differentiate :

$\bf{(x^{2} + 3x)^{2}}$

To find :

The Differentiation of $\bf{(x^{2} + 3x)^{2}}$.

Solution :

First let us solve the given equation [i.e, (x² + 3x)²]

$:\implies \bf{(x^{2} + 3x)^{2}}$

By using the identity , (a + b)² = a² + 2ab + b² , we get :

$:\implies \bf{(x^{2})^{2} + 2 \times x^{2} \times 3x + (3x)^{2}}$

$:\implies \bf{x^{4} + 6x^{3} + 9x^{2}}$

$\boxed{\therefore \bf{x^{4} + 6x^{3} + 9x^{2}}}$

Now by using the rule of Differentiation of Derivatives by dy/dx rule , we get :

$:\implies \bf{\dfrac{dy}{dx} = \dfrac{d(x^{4} + 6x^{3} + 9x^{2})}{dx}}$

$:\implies \bf{\dfrac{dy}{dx} = \dfrac{d(x^{4})}{dx} + \dfrac{d(6x^{3})}{dx} + \dfrac{d(9x^{2})}{dx}}$

$\boxed{\therefore \bf{\dfrac{dy}{dx} = \dfrac{d(x^{4})}{dx} + \dfrac{d(6x^{3})}{dx} + \dfrac{d(9x^{2})}{dx}}}$

Now let's Differentiate each single terms :

Differentiation of x⁴ :

We know the first principal of Differentiation i.e,

$\boxed{\bf{f'x = \lim_{h\:\to\:0} \dfrac{f(x + h) - f(x)}{h}}}$

Using the above equation and substituting the values in it , we get :

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{(x + h)^{4} - x^{4}}{h}}$

By using the binomial theorem , (a + b)⁴ = a⁴ + b⁴ + 4a³b + 4ab³ + 6x²h² , we get :

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{x^{4} + h^{4} + 4x^{3}h + 4xh^{3} + 6x^{2}h^{2} - x^{4}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{\not{x^{4}} + h^{4} + 4x^{3}h + 4xh^{3} + 6x^{2}h^{2} - x^{4}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{\not{x^{4}} + h^{4} + 4x^{3}h + 4xh^{3} + 6x^{2}h^{2} - \not{x^{4}}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{h^{4} + 4x^{3}h + 4xh^{3} + 6x^{2}h^{2}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{h^{4}}{h} + \dfrac{4x^{3}h}{h} + \dfrac{4xh^{3}}{h} + \dfrac{6x^{2}h^{2}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} h^{3} + 4x^{3} + 4xh^{2} + 6x^{2}h^{1}}$

$:\implies \bf{f'(x) = 0^{3} + 4x^{3} + 4x(0)^{2} + 6x^{2}(0)}$

$:\implies \bf{f'(x) = 4x^{3}}$

$\boxed{\therefore \bf{f'(x) = 4x^{3}}}$

Hence the Differentiation of x⁴ is 4x³.

Differentiation of 6x³ :

We know the first principal of Differentiation i.e,

$\boxed{\bf{f'x = \lim_{h\:\to\:0} \dfrac{f(x + h) - f(x)}{h}}}$

Using the above equation and substituting the values in it , we get :

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{6(x + h)^{3} - 6x^{3}}{h}}$

By using the identity , (a + b)³ = a³ + b³ + 3a³b + 3ab³, we get :

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{6(x^{3} + h^{3} + 3x^{2}h + 3xh^{2}) - 6x^{3}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{6x^{3} + 6h^{3} + 18x^{2}h + 18xh^{2} - 6x^{3}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{6h^{3} + 18x^{2}h + 18xh^{2}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{6h^{3}}{h} + \dfrac{18x^{2}h}{h} + \dfrac{18xh^{2}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} 6h^{2} + 18x^{2} + 18xh^{2}}$

$:\implies \bf{f'(x) = 6(0)^{2} + 18x^{2} + 18x(0)^{2}}$

$:\implies \bf{f'(x) = 18x^{2}}$

$\boxed{\therefore \bf{f'(x) = 18x^{2}}}$

Hence the Differentiation of 6x³ is 18x² .

Differentiation of 9x² :

We know the first principal of Differentiation i.e,

$\boxed{\bf{f'x = \lim_{h\:\to\:0} \dfrac{f(x + h) - f(x)}{h}}}$

Using the above equation and substituting the values in it , we get :

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{9(x + h)^{2} - 9x^{2}}{h}}$

By using the identity , (a + b)² = a² + 2ab + b² , we get :

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{9(x^{2} + 2xh + h^{2}) - 9x^{2}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{9x^{2} + 18xh + 9h^{2} - 9x^{2}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{18xh + 9h^{2}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} \dfrac{18xh}{h} + \dfrac{9h^{2}}{h}}$

$:\implies \bf{f'(x) = \lim_{h\:\to\:0} 18x + 9h}$

$:\implies \bf{f'(x) =18x + 9(0)}$

$:\implies \bf{f'(x) = 18x}$

$\boxed{\therefore \bf{f'(x) = 18x}}$

Hence the Differentiation of 9x² is 18x .

Now substituting them in the Equation :

$\boxed{\therefore \bf{\dfrac{dy}{dx} = \dfrac{d(x^{4})}{dx} + \dfrac{d(6x^{3})}{dx} + \dfrac{d(9x^{2})}{dx}}}$

We get ,

$:\implies \bf{\dfrac{dy}{dx} = 4x^{3} + 18x^{2} + 18x}$

$:\implies \bf{\dfrac{dy}{dx} = 4x^{3} + 18x^{2} + 18x}$

$\boxed{\therefore \bf{\dfrac{dy}{dx} = 4x^{3} + 18x^{2} + 18x}}$

Hence the Differentiation of (x² + 3x) is 4x³ + 28x² + 18.

Answer 2

Answer:

$\sf{\dfrac{dy}{dx}=\dfrac{d}{dx}(x^{2}+3x)^{2}}$

$\sf{\therefore{\dfrac{dy}{dx}=2(x^{2}+3x)\times\dfrac{d}{dx}(x^{2}+3x)}}$

$\sf{\therefore{\dfrac{dy}{dx}=2(x^{2}+3x)(2x+3)}}$

$\sf{\therefore{\dfrac{dy}{dx}=2(2x^{3}+9x^{2}+9x)}}$

$\sf{\therefore{\dfrac{dy}{dx}=4x^{3}+18x^{2}+18x}}$