The derivative of the natural log is 1/x, therefore the derivative is 1/cos(x). However, since the value of cos(x) is submitted within the natural log we must use the chain rule. Then, we multiply 1/cos(x) by the derivative of cos(x). We get the answer: -sin(x)/cos(x) which can be simplified into -tan(x).
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The integral of tan(x) dx = ln | sec(x) | + cto solve... tan(x) = sin(x)/cos(x)the integral of (sin(x)/cos(x) dx) ... let u = cos(x) then du = -sin(x) dx= the integral of (1/u -du)= -ln | u | + c= -ln | cos(x) | + c= ln | (cos(x))^-1 | + c ... or ... ln | 1/cos(x) | + c= ln | sec(x) | + c
The derivative of ln(10) is 1/10. This is because the derivative of the natural logarithm function ln(x) is 1/x. Therefore, when differentiating ln(10), the derivative is 1/10.
y = e^ln x using the fact that e to the ln x is just x, and the derivative of x is 1: y = x y' = 1
If the function is (ln x)2, then the chain rules gives us the derivative 2ln(x)/x, with the x in the denominator. If the function is ln (x2), then the chain rule gives us the derivative 2/x.
The derivative of 3cos(x) is -3sin(x). This can be found using the chain rule, which states that the derivative of a composition of functions is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In this case, the derivative of cos(x) is -sin(x), and when multiplied by the constant 3, we get -3sin(x) as the derivative of 3cos(x).