Question #0dc0b

3 Answers
Mar 25, 2017

#d/dx(sqrt(2x+3))=1/sqrt(2x+3)#

Explanation:

using #dx^n/dx=nx^(n-1)#.
and chain rule is: #dy/dx=dy/(du) * (du)/(dt) * (dt)/(dx)#
so, #d(sqrt(2x+3))/dx=d(sqrt(2x+3))/(d(2x+3)) * d(2x+3)/(dx) =1/2*(2x+3)^(1/2-1) * (2*1+0)=1/(sqrt(2x+3))#

Mar 25, 2017

I guessed that you meant #y = sqrt(2x+3)#.

Explanation:

The definition of a derivative is:

#y'(x) = lim_(hto0)(y(x+h)-y(x))/h#

Given:

#y(x) = sqrt(2x+3)#

Then:

#y(x+h) = sqrt(2(x+h)+3)#

Let's find an equation for h.

Square both sides:

#(y(x+h))^2 = 2(x+h)+3#

Use the distributive property:

#(y(x+h))^2 = 2x+ 2h+3#

Please observe that #2x + 3# is the same as #(y(x))^2#:

#(y(x+h))^2 = (y(x))^2 + 2h#

Subtract #(y(x))^2# from both side and flip the equation:

#2h = (y(x+h))^2 - (y(x))^2#

Divide both sides by 2:

#h = ((y(x+h))^2 - (y(x))^2)/2#

Substitute this for h into the definition:

#y'(x) = lim_(hto0)(y(x+h)-y(x))/(((y(x+h))^2 - (y(x))^2)/2)#

Flip the 2 up to the numerartor:

#y'(x) = lim_(hto0)(2(y(x+h)-y(x)))/((y(x+h))^2 - (y(x))^2)#

We know how the difference of two squares factors, #(a^2-b^2)=(a+b)(a-b)#, so we can do this to the denominator:

#y'(x) = lim_(hto0)(2(y(x+h)-y(x)))/((y(x+h) + y(x))(y(x+h) - y(x)))#

Please observe that a common factor cancels:

#y'(x) = lim_(hto0)(2cancel((y(x+h)-y(x))))/((y(x+h) + y(x))cancel((y(x+h) - y(x))))#

Here is the equation with the cancelled factors removed:

#y'(x) = lim_(hto0)2/(y(x+h) + y(x))#

Now it is ok to let #hto0#

#y'(x) = 2/(y(x) + y(x))#

#y'(x) = 2/(2y(x))#

#y'(x) = 1/(y(x))#

#y'(x) = 1/sqrt(2x+3)#

Mar 26, 2017

#1/sqrt(2x+3)#

Explanation:

#y= f(x)= sqrt(2x+3)#

#y'= f' (x)= lim_( h->0) (f(x+h) -f(x))/h#

=#lim_(h->0) (sqrt (2(x+h)+3) -sqrt(2x+3))/h#

=#lim_(h->0) (sqrt(2(x+h)+3) -sqrt(2x+3))/h * (sqrt(2(x+h) +3) +sqrt(2x+3))/(sqrt(2(x+h)+3) +sqrt(2x+3))#

=#lim_(h->0) ((2(x+h)+3)-(2x+3))/ h *1/(sqrt(2(x+h) +3) +sqrt(2x+3))#

=#lim_(h->0) (2h)/h * 1/(sqrt(2(x+h) +3)+sqrt(2x+3))#

=#2/(2sqrt(2x+3)#= #1/sqrt(2x+3)#