8f(x)+6f(1/x)=x+5 & y=x^2f(x) , find dy/dx at x=-1

Question

This is about calculus....... Can someone help me out in this

Answer 1

First plug -1 in for x in the first equation:

 

8f(x) + 6f(1/x) = x + 5

8f(-1) + 6f(1/(-1)) = -1 + 5

8f(-1) + 6f(-1) = 4

14f(-1) = 4

f(-1) = 4/14 = 2/7

 

Now use implicit differentiation on the first equation.  Note that we have to use the chain rule for the second term on the left

 

8f(x) + 6f(1/x) = x + 5

 

8f'(x) + 6f'(1/x) (-1/x²) = 1

 

Again, plug -1 in for x:

 

8f'(-1) + 6f'(1/-1) (-1/(-1)²) = 1

8f'(-1) + 6f'(-1) (-1/1) = 1

8f'(-1) + 6f'(-1) (-1) = 1

8f'(-1) - 6f'(-1) = 1

2f'(-1) = 1

f'(-1) = 1/2

 

Now we know the values of f(x) and f'(x) when x = -1.

Now differentiate the second equation:

 

y=x² f(x)

dy/dx = 2x f(x) + x² f'(x)

 

Plug in -1 for x:

 

dy/dx (when x = -1) = 2(-1)f'(-1) + (-1)² f(-1) = 

-2 (1/2) + 1 (2/7)

-1 + 2/7 =

-5/7

 

Recheck all my calculations to be sure, but that's the way to solve this.