00:01
Given the function y equal 9 plus 4x raised to the 4 over x, we want to use logarithmic differentiation to find the derivative of y respect to x in par a.
00:19
In par b, we find the slope of the tiny line at x equal 1, and per c we find the equation of 10 in line at x equal 1.
00:28
So the idea of logarithmic differentiation is to apply logarithm, both sides, of the definition of the equation of the function.
00:42
And so we apply natural logarithm, for example here.
00:46
We get natural logarithm of y is natural logarithm of this power here.
00:52
And we apply properties of the lower rhythms.
00:56
In this case, we know the logarithm of the power is equal to the exponent times the logarithm of the base.
01:06
So we have this formula here.
01:10
Then, from here we can differentiate with respect to x both sides of the equation.
01:18
And so on the left we get by applying the same rule derivative of y, that is, let's put better derivative of y with respect to x like this and that divided by y.
01:33
It is because the derivative of natural algorithm of y is 1 over y times by applying the same rule times the derivative of y respect to x and we get this fraction here.
01:46
So that must be equal to the derivative respect to x.
01:51
Let me put this equation first here.
01:55
Relative respect to x of natural logarithm of y will be derivative respect to x of this expression over here, 4 over x times natural algorithm of 9 plus 4x.
02:10
Let me move this little bit to the left and all this also so we get this let me put a parenthesis that remains open that's this one that's what we get derivative respect to x of let me get rid of this because we have it already here so we get derivative of a product of two functions so we get derivative the first function derivative of four over x is now a negative 4 over x square times natural logarithm of 9 plus 4x plus 4x plus 4x derivative of natural algorithm of 9 plus 4x is 1 over 9 plus 4x times the derivative of 9 plus 4x applying the shen rule and we get 4 over 4 which is derivative of 9 plus 4x respect to x over 9 plus 4x so we have this expression here and we get by solving for the derivative of y respect to x we get that that derivative is y times this expression here we can get out a common factor let's say 4 over x so 4 y over x let me this a little bit better is 4.
04:02
This 4 over x here the y we have up here and then that over x times and then we get this term here which is positive we get 4 over 9 plus 4x minus natural logarithm of 9 plus 4x divided by x good so we have that and as you can see we have the function here so we get to put the expression of the function right there to say that the derivative of y respect to x is 4 times 9 plus 4x to the 4 over x right over x all that times 4 over 9 plus 4x minus natural logarithm of 9 plus 4x over x and that we can say is expression for the derivative of the function the given function y respect to and as you can see is way easier to do this...