How do you differentiate # f(x)= (9-x)sqrt(x^2 -4) # using the product rule?

Application of the product rule shows that this function's derivative is equal to

#f'(x)=sqrt(x^2-4)d/dx(9-x)+(9-x)d/dxsqrt(x^2-4)#

The product rule basically states that the derivative of two functions, when multiplied by one another, is equal to the derivative of one function multiplied by the other, nondifferentiated, plus the version of this with the roles switched (the previously differentiated function now in its original form, with the previously original function now differentiated).

First, let's find each of the derivatives the application of the product rule spawned outside of the original expression.

#d/dx(9-x)=-1#

Since the coefficient on the #x# term is #-1# (this is deceiving).

To differentiate the square root function, we'll need the chain rule:

#d/dx(g(x))^(1/2)=1/2(g(x))^(-1/2)*g'(x)=(g'(x))/(2sqrt(g(x)))#

Here, the function "inside" of the square root is #g(x)=x^2-4#, so

#d/dxsqrt(x^2-4)=(d/dx(x^2-4))/(2sqrt(x^2-4))=(2x)/(2sqrt(x^2-4))=x/sqrt(x^2-4)#

Plug both of these derivatives back into the original equation:

#f'(x)=sqrt(x^2-4)(-1)+(9-x)(x/sqrt(x^2-4))#

Now, this just becomes a matter of simplification.

#f'(x)=-sqrt(x^2-4)+(x(9-x))/sqrt(x^2-4)#

Multiply the first term by #sqrt(x^2-4)/sqrt(x^2-4)# so there can be a common denominator. Also, note that this will cause some nice simplification, since #(sqrt(x^2-4))^2=x^2-4#.

#f'(x)=(-(x^2-4)+x(9-x))/sqrt(x^2-4)#

#f'(x)=(-2x^2+9x+4)/sqrt(x^2-4)#