What is the derivative of the function ? y = x √ 2x + 3

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What is the derivative of the function ? y = x √ 2x + 3

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Answer 1 · 2018-03-08T16:10:23

done it for you

Explanation:

Here I have used the uv rule for differentiation Kevin's MAth

Answer 2 · 2018-03-08T16:18:43

solution of $y = x \sqrt{2 x + 3}$ $y=xsqrt(2x+3)$ instead of $y = x \sqrt{2} x + 3$ $y=xsqrt(2)x+3$
$\frac{d y}{d x} = \frac{3 (x + 1)}{\sqrt{2 x + 3}}$ $(dy)/(dx)=(3(x+1))/sqrt(2x+3)$
HINT:
$(1) y = u \cdot v \Rightarrow \frac{d y}{d x} = u \cdot \frac{d v}{d x} + v \cdot \frac{d u}{d x}$ $color(red)((1)y=u*v=>(dy)/(dx)=u*(dv)/(dx)+v*(du)/(dx)$

Explanation:

$y = x \cdot \sqrt{2 x + 3} = x \cdot {(2 x + 3)}^{\frac{1}{2}}$ $y=x*sqrt(2x+3)=x*(2x+3)^(1/2)$ $\Rightarrow \frac{d y}{d x} = x \cdot \frac{d}{d x} ({(2 x + 3)}^{\frac{1}{2}}) + {(2 x + 3)}^{\frac{1}{2}} \cdot \frac{d}{d x} (x)$ $=>(dy)/(dx)=x*d/(dx)((2x+3)^(1/2))+(2x+3)^(1/2)*d/(dx)(x)$
$=x*1/cancel(2)*(2x+3)^(1/2-1)*cancel(2)+(2x+3)^(1/2)$
$=x(2x+3)^(-1/2)+(2x+3)^(1/2$
$=x/sqrt(2x+3)+sqrt(2x+3)$
$=(x+2x+3)/(sqrt(2x+3))=(3x+3)/sqrt(2x+3)=(3(x+1))/sqrt(2x+3)$
We can multiply $x$ with $sqrt(2x+3)$
$y=x(2x+3)^(1/2)=(x^2)^(1/2)(2x+3)^(1/2)=[x^2(2x+3)]^(1/2)$ $=>y=(2x^3+3x^2)^(1/2)=>y^'=1/2(2x^3+3x^2)^(1-1/2)(6x^2+6x)$
$y^'=(6x)/2(2x^3+3x^2)^(-1/2)(x+1)=(3x(x+1))/sqrt(2x^3+3x^2)=(3x(x+1))/(sqrt(x^2(2x+3)))=(3x(x+1))/(xsqrt((2x+3)))=(3(x+1))/sqrt(2x+3)$
Note:: $tox*sqrt(2x+3)!=x+sqrt(2x+3)$

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What is the derivative of the function ? y = x √ 2x + 3

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