I find that problems like this are solved easier by applying our rules of logarithmic functions. #y = x*sqrt(16-x^2)# We know from the rules of algebra that manipulated one side of an equation by applying an operator to both sides is still an equivalent function, but in an alternative form. What I mean by this, #y = x*sqrt(16-x^2) = ln(y) = ln(x*sqrt(16-x^2))# In essences we have not changed the function, but added #0#.
Going back to our log rules we know that #ln(x*y) = ln(x) + ln(y)# , so #ln(y) = ln(x*sqrt(16-x^2)) = ln(y) = ln(x) + ln(sqrt(16-x^2))# .
From here, let us find #d/dx ln(y) = ln(x) + ln(sqrt(16-x^2))# . Using implicit differentiation, we find the derivative of both sides of our equation. #d/dx ln(y) =(1/y)(dy/dx)# where we find the derivative of the functions operator, #d/dx ln(y) = 1/y# then multiple the derivative by the interior function #d/dx y = dy/dx# .
From here we find the derivative of the other side of our function.
#d/dx ln(x) + ln(sqrt(16-x^2))# we know from our rules of derivatives that these two derivative can be split and later added together finding the derivative of the first function then the second. #d/dx ln(x) = 1/x#
before we continue finding the derivative of #ln(sqrt(16-x^2))# let us further note that this function is equivalent to #ln((16-x^2))^(1/2)# . Note that our rules of logs state that we can move the power of a log in from of the function. #ln(16-x^2)^(1/2) = 1/2ln(16-x^2)# . Now let's find the derivative #d/dx1/2ln(16-x^2)# where #1/2 = c# a constant and remains derivative until the end. #d/dx ln(16-x^2) = (1/2)(1/(16-x^2))(-2x)# the #1/2# simplifies #-2x = -x = ((-x)/(16-x^2))#
Now let us add these functions together, #(1/y)(dy/dx) = 1/x-((x)/(16-x^2))# . We cannot leave this function in this form though. Multipling both sides of our function by #y# noting #y = x*sqrt(16-x^2)# returns #dy/dx = (x*sqrt(16-x^2))(1/x-((x)/(16-x^2))# our final answer.
