DIfferentiation

Question

``Given that dy/dx=(a/x^2)+1 and that when x=1 dy/dx=3 and y=3 find the value of y when x=2.''.

Answer 1

Set the derivative equal to each other.

 

(a/x²)+ 1 = 3

 

 

Then plug in x=1 into this equation to solve for a.

 

(a/1) + 1 = 3

 

a/1 = 2

 

a = 2

 

 

So your derivative is

 

dy/dx = 2x^-2 + 1

 

 

Then take the anti-derivative to find y in terms of x.

 

y = 2(x^-1 / -1) + x + C

 

y = (-2 / x) + x + C

 

 

Then set y=3 and x=1 to solve for C.

 

3 = (-2 / 1) + 1 + C

 

3 = -2 + 1 + C

 

3 = -1 + C

 

4 = C

 

 

 

Now evaluate y when x=2.

 

y = (-2 / 2) + 2 + 4

 

y = -1 + 6

 

y = 5

Answer 2

dy/dx = a/x² + 1

 

dy/dx = 3 when x =1 and y = 3

 

3 = a/1 + 1

3 = a + 1

a = 2

 

dy/dx = 2/x² + 1

 

∫ dy = ∫ 2/x² + 1 dx

 

 y + c  = -2/x + x + c

 

combine unknown constants into one

 

y = -2/x + x + c

 

y = 3 when x =1

 

3 = -2/1 + 1 + c

 

3 = -1 + c

 

c = 4

 

y = -2/x + x + 4

 

check: √   dy/dx = 2/x² + 1

 

Now when x = 2

 

y = -2/2 + 2 + 4

 

y = -1 + 6

 

y = 5