Question

Differentiate y=√[x^2+4x-3]?

Answer 1

`dy/dx=(x+2)/(x^2+4x+3)^(1/2)`

Explanation:

`"differentiate using the " color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larr" chain rule"`

`y=sqrt(x^2+4x-3)=(x^2+4x-3)^(1/2)`

`rArrdy/dx=1/2(x^2+4x-3)^(-1/2)xxd/dx(x^2+4x-3)`

`color(white)(rArrdy/dx)=(2(x+2))/(2(x^2+4x-3)^(1/2))`

`color(white)(rArrdy/dx)=(x+2)/(x^2+4x-3)^(1/2)`

Answer 2

`(2x+4)/(2sqrt(x^2+4x-3)) = (x+2)/sqrt(x^2+4x-3)`

Explanation:

The chain rule states that given a composition of functions `f(x) = g(h(x)), (df)/dx = (dh)/dx * (dg)/(dh)`. Essentially, for the second part, we treat h as though it were a variable, and differentiate g with respect to that variable.

In this case, our function `h(x) = x^2+4x-3`, and `g(h)=sqrt h`.

From the power rule, we can determine `(dh)/dx = 2x + 4`. We can do the same for `g(h) = sqrt h = h^(1/2) - > (dg)/(dh) = h^(-1/2)/2 = 1/(2sqrth)`. This gives us:

`(df)/dx = (2x+4)(1/(2sqrth))`

Substituting `x^2+4x-3` back in for h, we get...

`(df)/dx = (2x+4)/(2sqrt(x^2+4x-3)) -> (x+2)/sqrt(x^2+4x-3)`