how do you find the Derivative of this problem?

Question

find the derivative of..


ds/dt if s= t/(2t+1)


answer: 1/(2t+1)² but i need to know how to get there

thanks!

Answer 1

your equation is equivalent to finding the derivative of t/(2t+1)

let f = t and g = (2t+1)
use the formula to find the derivative of f/g

= (f'g-fg')/g^2
f' =1 g'= 2
(f'g -fg')/g^2=
(1(2t+1)-2t)/(2t+1)^2 =
Answer is : 1/(2t+1)^2

Answer 2

Product rule:

d(fg) = f*dg + g*df

Let f = t, then df = 1
Let g = (2t + 1)^-1
then dg = 2*(-1) * (2t+1)^-2
...the 2 is from the 2t, the -1 is from the exponent before deriving, and the ^-2 is from the old exponent, minus 1.

Then s = fg

ds = t[(2)*(-1)*(2t+1)^-2] + (2t+1)^-1 * (1)
ds = -2t / (2t+1)^2 + 1 / (2t+1)

put both terms on a common denominator (2t+1)^2

ds = -2t / (2t+1)^2 + (2t+1) / (2t+1)^2
ds = (-2t + 2t + 1) / (2t+1)^2

ds = what you got.

Answer 3

Use the quotient rule for the derivation:

if s(t)=t/(2t+1) = u(t)/v(t)
u(t)=t
v(t)=(2t+1)

then is ds/dt = [u`(t)*v(t) - u(t)*v`(t)] / [v(t)]^2

u`(t)=1
v`(t)=2
[v(t)]^2=(2t+1)^2

===>ds/dt=1/(2t+1)^2

Answer 4

let t is f and 2t+t is g, the formula, (gf' - fg')/g^2

so [(2t+1) * d/dx (t) - t * d/dx (2t+1)] / (2t+1) ^2
=[(2t+1) * 1 - t * 2] / (2t+1) ^2 ,<----because d/dx (t) =1, and d/dx (2t)=2. d/dx (1)=0, so d/dx (2t+1)= 2+0=2

=[2t+1-2t ] / (2t+1) ^2
=1 / (2t+1) ^2