I need help on calc 2

Question

derivative of 7^root(x-3)

Answer 1

Hi Fred

 

The derivative of the ratio of two functions (quotient rule)  is:

 

d/dx (f/g) = ((df/dx)xg - fx(dg/dx) )/g²

                  

 

Here   f(x) = 7      df/dx = 0

 

          g(x) = root(x-3)    But g(x) is a composed function.

 

          dg/dx = (1/2) (x-3)^-3/2(1)   because the derivative of x-3 is 1

 

Then,

 

d/dx (f/g) = ( 0*g - 7x(1/2)x(x - 3)^-3/2) / (root(x-3))²

 

d/dx (f/g) = ( -(7/2)(x - 3)^-3/2) / (x -3)

 

d/dx (f/g) = (-7/2)*(X -3)^-3/2-1  = (-7/2)(X -3)^-5/2

 

I hope it helped you

 

 

Answer 2

y = 7^√(x-3)

 

To find dy/dx, I suggest using the logarithmic derivative.  I will give it a try:

 

Take ln() on both sides and get

 

ln(y) = √(x-3) ln7

 

Take d/dx on both sides and get

 

(1/y) dy/dx = ln7 (1/2) / √(x-3)

 

Multiply both sides by y and get

 

dy/dx = y ln7 (1/2) / √(x-3) = 7^√(x-3) ln7 (1/2) / √(x-3) 

 

dy/dx = (ln7 / 2) 7^√(x-3) / √(x-3)