00:01
We are going to use logarithmic differentiation to find derivative of y respect to x when y is equal to cosine square of x times e to the x squared minus x times x equals 1 to the 2 thirds.
00:19
So we use generally logarithmic differentiation where we have a function defined as a product of several functions like in this case.
00:29
Function y, function of x, is the product of three functions, this one here, this one here, and this one here.
00:39
So instead of trying to apply derivative directly by using the derivative of the product, we can use logarithmic differentiation, which consists of applied the natural logarithm to y and then find the derivative that way.
00:59
So natural logarithm of y is equal to natural logarithm of this product of the three functions.
01:15
And the key thing is that the logarithm of a product is a sum of the logarithms.
01:23
So we know that this is the logarithm of cosine square of x plus the natural logarithm of e to the x squared minus x plus the natural logarithm of e to the x squared minus x plus an of x plus 1 to the 2 thirds.
01:45
And now we apply the property of the logarithm telling us that the logarithm of power is the exponent times the logarithm of the base.
01:59
So in this case we have cosine of x squared, and so this is 2 times, it's 1 in 2 here, the natural logarithm of the base cosine of x.
02:11
Plus and here we are composing the natural logarithm and the exponential function which are inverse to each other so the result is the exponent x squared minus x plus and here we apply again the property of the logarithm telling us that the natural algorithm of a power is equal the exponent times exponents here is two thirds times the natural algorithm of the base x plus 1 and so this is the natural algorithm of y.
02:49
That is, natural the algorithm of y is two natural logarithm of cosine of x plus x squared minus x plus two thirds, natural logarithm of x plus y.
03:06
And now we differentiate with respect to x both sides.
03:12
So the derivative respect to x of the natural logarithm of y is equal to the derivative respect to x of two natural logarithm of cosine of x plus x squared minus x plus two thirds natural algorithm of x plus 1.
03:36
So here we know is 1 over y times derivative of y respect to x applying the same rule because y depends on x and that's equal to 2 is a constant so the derivative of natural algorithm of cosine of x is 1 over cosine of x times derivative respect to x of cosine of x applying the same rule here plus 2x minus 1 plus 2 3rds times and here we get 1 over x plus 1 it should put here as a valid but we're going to get rid of that for the time being times derivative respect to x of x plus 1 so i put some space here put it to the left a little bit okay derivative respect to f x of x plus 1 and so 1 over y times derivative of y respect to x is 2 over cosine of x times derivative of cosine of x is negative sign of x plus 2x minus 1 plus 2 over 3 times x plus 1 and this derivative is equal to 1 so we get that and negative sine of x divided by cosine of x is negative tangent of x so we get here 1 of y times the derivative of y respect to x equal negative 2 tangent of x plus 2x minus 1 plus 2 over 3 times x plus 1 and so derivative of y respect to x which is the same as y derivative written this way is y times negative 2 tangent of x plus 2 x minus 1 plus 2 over 3 times x plus 1 because this y here pass to the right multiplying the expression on the right.
06:30
And now we put the expression of y if we want...