Question

find derivative of this..............​

Answer 1

Question :

$\sf{If\:f(x) = \dfrac{2x^{3}k(x) - 3}{3x + 2}\:then\:find\:\dfrac{d[f(x)]}{dx}.}$

Solution :

$\sf{Given,\:the\:function\:of\:f(x)\:is\:\dfrac{2x^{3}k(x) - 3}{3x + 2}}\:$$\sf{which\:can\:also\:be\:written\:as\:\dfrac{2x^{4}k - 3}{3x + 2}}$

$\textsf{Now, by using the quotient rule of differentiation}$

$\textsf{and differentiating the function of f(x) with respect to x, we get :}$

$\underline{\sf{Quotient\:rule\:of\:differentiation\::- }}\\ \\ \sf{\dfrac{d}{dx}\bigg(\dfrac{u}{v}\bigg) = \dfrac{(v)\dfrac{d(u)}{dx} - (u)\dfrac{d(v)}{dx}}{(v)^{2}}}$

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

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

$\textsf{Now, by using the power rule of differentiationand}$$\textsf{constant rule of differentiation, we get :}\\ \\ \bullet \underline{\sf{Power\:rule\:of\:differentiation\::- }}\\ \\ \sf{\dfrac{d(x^{n})}{dx} = n \cdot x^{(n - 1)}} \\ \\ \bullet \underline{\sf{Constant\:rule\:of\:differentiation\::- }}\\ \\ \sf{\dfrac{d(c)}{dx} = 0}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{2x^{4}k - 3}{3x + 2}\bigg) = \dfrac{(3x + 2)(8x^{3}k) - 0 - (2x^{4}k - 3)3 + 0}{(3x + 2)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{2x^{4}k - 3}{3x + 2}\bigg) = \dfrac{(3x + 2)(8x^{3}k) - (2x^{4}k - 3)3}{(3x + 2)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{2x^{4}k - 3}{3x + 2}\bigg) = \dfrac{24x^{4}k + 16x^{3}k - 6x^{4}k + 9}{(3x + 2)^{2}}}$

$:\implies \sf{\dfrac{d}{dx}\bigg(\dfrac{2x^{4}k - 3}{3x + 2}\bigg) = \dfrac{18x^{4}k + 16x^{3}k + 9}{(3x + 2)^{2}}}$

$\boxed{\therefore \sf{\dfrac{d}{dx}\bigg(\dfrac{2x^{4}k - 3}{3x + 2}\bigg) = \dfrac{18x^{4}k + 16x^{3}k + 9}{(3x + 2)^{2}}}}$