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)