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

2 Answers
Jeremiah Kevin M.
Mar 8, 2018

done it for you

Explanation:

Here I have used the uv rule for differentiationKevin's MAth

maganbhai P.
Mar 8, 2018

solution of #y=xsqrt(2x+3)# instead of #y=xsqrt(2)x+3#
#(dy)/(dx)=(3(x+1))/sqrt(2x+3)#
HINT:
#color(red)((1)y=u*v=>(dy)/(dx)=u*(dv)/(dx)+v*(du)/(dx)#

Explanation:

#y=x*sqrt(2x+3)=x*(2x+3)^(1/2)##=>(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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