Question

Hola brainly wizards,

Question:-

Differentiate w.r.t.x: (x + 1)² + (x + 2)³. (x + 3)⁴.

Plz solve it's urgent.​

Answer 1

Solution :

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(x + 1)^{2} + (x + 2)^{3} \cdot (x + 3)^{4}]}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(x + 1)^{2}]}{dx} + \dfrac{d[(x + 2)^{3} \cdot (x + 3)^{4}]}{dx}}$

$\textsf{Now, let us differentiate}\: \sf{(x + 2)^{3} \cdot (x + 3)^{4}} \textsf{of the equation :}$

$\underline{\sf{Product\:rule\:of \:differentiation :-}} \\ \\ \sf{\dfrac{d(uv)}{dx} = (u)\cdot \dfrac{d(v)}{dx} + (v) \cdot \dfrac{d(u)}{dx}}$

$\textsf{By using the product rule of differentiation, and substituting the values in it, we get :}$

$:\implies \sf{\dfrac{d(uv)}{dx} = [(x + 2)^{3}] \cdot \dfrac{d[(x + 3)^{4}]}{dx} + [(x + 3)^{4}] \cdot \dfrac{d[(x + 2)^{3}]}{dx}}$

$\textsf{By using the power rule of differentiation, we get :}$

$\underline{\sf{Power\:rule\:of \:differentiation :-}} \\ \\ \sf{\dfrac{d(x^{n})}{dx} = n\cdot x^{(n - 1)}}$

$:\implies \sf{\dfrac{d(uv)}{dx} = (x + 2)^{3} \cdot 4(x + 3)^{(4 - 1)} + (x + 3)^{4} \cdot 3(x + 2)^{(3 - 1)}}$

$:\implies \sf{\dfrac{d(uv)}{dx} = (x + 2)^{3} \cdot 4(x + 3)^{3} + (x + 3)^{4} \cdot 3(x + 2)^{2}}$

$:\implies \sf{\dfrac{d(uv)}{dx} = (x + 2)^{3} \cdot 4(x + 3)^{3} + (x + 3)^{4} \cdot 3(x + 2)^{2}}$

$\boxed{\therefore \sf{\dfrac{d(uv)}{dx} = (x + 2)^{3} \cdot 4(x + 3)^{3} + (x + 3)^{4} \cdot 3(x + 2)^{2}}}$

$\textsf{Now, let us differentiate}\: \sf{(x + 1)^{2}} \textsf{of the equation :}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(x + 1)^{2}]}{dx}}$

$\textsf{By using the power rule of differentiation, we get :}$

$\underline{\sf{Power\:rule\:of \:differentiation :-}} \\ \\ \sf{\dfrac{d(x^{n})}{dx} = n\cdot x^{(n - 1)}}$

$:\implies \sf{\dfrac{dy}{dx} = 2(x + 1)^{(2 - 1)}}$

$:\implies \sf{\dfrac{dy}{dx} = 2(x + 1)^{1}}$

$\boxed{\therefore \sf{\dfrac{dy}{dx} = 2(x + 1)}}$

$\textsf{Now, by substituting the derivative of}\: \sf{(x + 1)^{2}} and \sf{(x + 2)^{3} \cdot (x + 3)^{4}} \textsf{in the equation, we get :}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[(x + 1)^{2}]}{dx} + \dfrac{d[(x + 2)^{3} \cdot (x + 3)^{4}]}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = 2(x + 1) + (x + 2)^{3} \cdot 4(x + 3)^{3} + (x + 3)^{4} \cdot 3(x + 2)^{2}}$

$\underline{\underline{\therefore \sf{\dfrac{dy}{dx} = 2(x + 1) + (x + 2)^{3} \cdot 4(x + 3)^{3} + (x + 3)^{4} \cdot 3(x + 2)^{2}}}}$

Answer 2

$\Huge\underline{\overline{\mid{\green{Answer}}\mid}}$

Refer to the attachment.

$\Huge{-------------}$

Formulas used :-

• log xⁿ = nlogx

• d/dx(log x) = 1/x

•d/dx(xⁿ) = nxⁿ-¹

×××Logarithmic Differentiation×××

$\Huge \boxed{ \colorbox{gold}{hope \: this \: helps}}$