How to differentiate $y = {(x + 2)}^{2} {(3 x - 1)}^{3}$ $y=(x+2)^2(3x-1)^3$ , find $\frac{d y}{d x}$ $dy/dx$ as a product of its factor ?

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Answer 1 · 2018-07-17T14:18:22

$\frac{d y}{d x} = (x + 2) (15 x + 16) {(3 x - 1)}^{2}$ $dy/dx=(x+2)(15x+16)(3x-1)^2$

Given that

$y = {(x + 2)}^{2} {(3 x - 1)}^{3}$ $y=(x+2)^2(3x-1)^3$

differentiating w.r.t. $x$ $x$ as follows

$\frac{d y}{d x} = \frac{d}{d x} ({(x + 2)}^{2} {(3 x - 1)}^{3})$ $dy/dx=d/dx((x+2)^2(3x-1)^3)$

$= {(x + 2)}^{2} \frac{d}{d x} {(3 x - 1)}^{3} + {(3 x - 1)}^{3} \frac{d}{d x} {(x + 2)}^{2}$ $=(x+2)^2d/dx(3x-1)^3+(3x-1)^3d/dx(x+2)^2$

$= {(x + 2)}^{2} (3 {(3 x - 1)}^{2} (3)) + {(3 x - 1)}^{3} (2 (x + 2))$ $=(x+2)^2(3(3x-1)^2(3))+(3x-1)^3(2(x+2))$

$= 9 {(x + 2)}^{2} {(3 x - 1)}^{2} + 2 (x + 2) {(3 x - 1)}^{3}$ $=9(x+2)^2(3x-1)^2+2(x+2)(3x-1)^3$

$= (x + 2) {(3 x - 1)}^{2} (9 x + 18 + 6 x - 2)$ $=(x+2)(3x-1)^2(9x+18+6x-2)$

$= (x + 2) {(3 x - 1)}^{2} (15 x + 16)$ $=(x+2)(3x-1)^2(15x+16)$

$= (x + 2) (15 x + 16) {(3 x - 1)}^{2}$ $=(x+2)(15x+16)(3x-1)^2$

Answer 2 · 2018-07-17T14:23:21

$\frac{d y}{d x} = 2 (x + 2) {(3 x - 1)}^{3} + 9 {(x + 2)}^{2} {(3 x - 1)}^{2}$ $dy/dx=2(x+2)(3x-1)^3+9(x+2)^2(3x-1)^2$

$(uv)'=u'v+uv'$

$(f(g(x)))'=f'(g(x))*g'(x)$ we get

$dy/dx=2(x+2)(3x-1)^3+(x+2)^2*3(3x-1)^2*3$
this is equal to

$dy/dx=2(x+2)(3x-1)^3+9(x+2)^2(3x-1)^2$

How to differentiate y=(x+2)^2(3x-1)^3y=(x+2)2(3x−1)3y=(x+2)^2(3x-1)^3 , find dy/dxdydxdy/dx as a product of its factor ?