using a table find the derivatives of a combined function. I keep getting it wrong.

Question

The table is x f(x) g(x) f'(x) g'(x) f''(x)

                  0  -2   1     -1     4     3

                  1  3     2     3     -2   -4

                  2   1     0    -2    3     1

                  3   2     1     4     2     2

so I'm stuck on parts c and d.

For part c the question is v(x)=(f(x))^2   what is v''(3)

so I tried 8, then I tired 16 but I keep getting it wrong and I'm suck on another way to figure this out.

 

Then part d says r(x) = g(x)/x^3   find r'(3) so I tied using the quotient rule and chain rule and I got 27/243 but I got this wrong as well.

Please help me sort out my mistakes, I would appreciate it so much.

Answer 1

c. v(x) = (f(x))²

 

   v'(x) = 2f(x)f'(x)   [By the chain rule for powers}

 

   v"(x) = 2[f'(x)f'(x) + f(x)f"(x)]  [By the product rule]

 

   v"(3) = 2[f'(3)f'(3) + f(3)f"(3)]

 

           = 2[(4)(4) + (2)(2)] = 40

 

d.  r(x) = g(x)/x³

 

    By the quotient rule, r'(x) = [x³g'(x) - 3x²g(x)]/x⁶

 

                                 r'(3) = [27g'(3) - 27g(3)]/729

 

                                        = [27(2) - 27(1)]/729

 

                                        = 27/729 = 1/27

 

Answer 2

Part c)

 

v(x) = [f(x)]²

 

 

Find the second derivative of both sides of the equation using chain rule.

 

v'(x) = 2f(x) * f'(x)

 

 

Take the derivative of this derivative using product rule.  Do this carefully.

 

v''(x) = 2f'(x)f'(x) + 2f(x)f''(x)

 

 

Plug in x=3 into this second derivative.

 

v''(3) = 2[f'(3)]² + 2f(3)f''(3)

 

v''(3) = 2(4)² + 2(2)(2)

 

v''(3) = 2(16) + 8

 

v''(3) = 40

 

 

Part d)

 

r(x) = g(x) / x³

 

 

Take the derivative of both sides of the equation.  Use quotient rule.

 

r'(x) = [x³g'(x) - 3x²g(x)] / x⁶

 

 

Now we plug in x=3.

 

r'(3) = [3³g'(3) - 3(3)²g(3)] / (3)⁶

 

 

Now just use the table to plug in the values and simplify like I showed you in part c.