Question

Question #0dc0b

Answer 1

`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))#

Answer 2

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)`

Answer 3

`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)`