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Answer 1

Solution :

$\begin{lgathered}:\implies [tex] \sf{\dfrac{dy}{dx}$ =$\dfrac{d[(x + 1)^{2} + (x + 2)^{3} \cdot (x + 3)^{4}]}{dx}} \\ \\\end{lgathered}$⟹dxdy=dxd[(x+1)2+(x+2)3⋅(x+3)4]

$[tex] \begin{lgathered}$ \implies \sf{\dfrac{dy}{dx}[/tex] =$\dfrac{d[(x + 1)^{2}]}{dx}$ + $\dfrac{d[(x + 2)^{3} \cdot (x + 3)^{4}]}{dx}}\\ \\\end{lgathered}$⟹dxdy=dxd[(x+1)2]+dxd[(x+2)3⋅(x+3)4]

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

$\begin{lgathered}\underline{\sf{Product\:rule\:of \:differentiation :-}} \\ \\ \sf{\dfrac{d(uv)}{dx} = (u)\cdot \dfrac{d(v)}{dx} + (v) \cdot \dfrac{d(u)}{dx}} \\ \\ \\\end{lgathered}$Product rule of differentiation:$−dxd(uv)=(u)⋅dxd(v)+(v)⋅dxd(u)$

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

$\begin{lgathered}:\implies \sf{\dfrac{d(uv)}{dx} = [(x + 2)^{3}] \cdot \dfrac{d[(x + 3)^{4}]}{dx} + [(x + 3)^{4}] \cdot \dfrac{d[(x + 2)^{3}]}{dx}} \\ \\\end{lgathered}$:⟹dxd(uv)=[(x+2)3]⋅dxd[(x+3)4]+[(x+3)4]⋅dxd[(x+2)3]

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

$\begin{lgathered}\underline{\sf{Power\:rule\:of \:differentiation :-}} \\ \\ \sf{\dfrac{d(x^{n})}{dx} = n\cdot x^{(n - 1)}} \\ \\ \\\end{lgathered}$Power rule of differentiation:−dxd(xn)=n⋅x(n−1)

$\begin{lgathered}:\implies \sf{\dfrac{d(uv)}{dx} = (x + 2)^{3} \cdot 4(x + 3)^{(4 - 1)} + (x + 3)^{4} \cdot 3(x + 2)^{(3 - 1)}} \\ \\\end{lgathered}$⟹dxd(uv)=(x+2)3⋅4(x+3)(4−1)+(x+3)4⋅3(x+2)(3−1)

$\begin{lgathered}:\implies \sf{\dfrac{d(uv)}{dx} = (x + 2)^{3} \cdot 4(x + 3)^{3} + (x + 3)^{4} \cdot 3(x + 2)^{2}} \\ \\\end{lgathered}$⟹dxd(uv)=(x+2)3⋅4(x+3)3+(x+3)4⋅3(x+2)2

$\begin{lgathered}:\implies \sf{\dfrac{d(uv)}{dx} = (x + 2)^{3} \cdot 4(x + 3)^{3} + (x + 3)^{4} \cdot 3(x + 2)^{2}} \\ \\\end{lgathered}$:⟹dxd(uv)=(x+2)3⋅4(x+3)3+(x+3)4⋅3(x+2)2

$\begin{lgathered}\boxed{\therefore \sf{\dfrac{d(uv)}{dx} = (x + 2)^{3} \cdot 4(x + 3)^{3} + (x + 3)^{4} \cdot 3(x + 2)^{2}}} \\ \\\end{lgathered}$∴dxd(uv)=(x+2)3⋅4(x+3)3+(x+3)4⋅3(x+2)2

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

Now, let us differentiate(x+1)