Question

help pls how to differentiate (2x+1)/2sqrt(x^2+x-2)

Answer 1

I'm going to assume that you mean the following:\[\frac{ 2x+1 }{ 2\sqrt(x^2+x-2) }\]

Answer 2

Since this is a quotient, apply the quotient rule for differentiation.

Answer 3

yep thats the problem

Answer 4

The formula for the derivative of the quotient y=u/v is\[\frac{ dy }{ dx }=\frac{ v \frac{ du }{ dx }-u \frac{ dv }{ dx } }{ v^2 }\]

Answer 5

Another approach you could consider, though not necessarily as simple as just using the quotient rule. Notice that the numerator is the derivative of the highlighted term in the denominator:
\[\frac{ 2x+1 }{ 2\sqrt{\color{red}{x^2+x-2}}}\]
So if you were to let \(t=x^2+x-2\), you'd have
\[\frac{t'}{ 2\sqrt t}=\frac{1}{2}t^{-1/2}t'\]

Answer 6

If u = 2x+1, what is du/dx ?

If v = 2 Sqrt(x^2+2-2), what is dv/dx ?

Answer 7

the approach suggested by sithsandgiggles is an excellent one for this particular problem.

Answer 8

u = 2 and v=(x^2+x-2)(2x+1) right?

Answer 9

Actually,
u=2x+1, so that du/dx = 2. (Label every quantity properly.)

v=2Sqrt(x^2+x-2). What is dv / dx ?

Answer 10

Note that the quantity x^2 + x - 2 has the derivative 2x + 1.

Thus, you could re-write \[\frac{ 2x+1 }{ 2\sqrt{\color{red}{x^2+x-2}}}\]