Differentiate the function. Simplify.?

Question

f(x) = cos (ln 10x)

Answer 1

f(x) = cos( ln(10x) )

I'm going to try a unique approach; the good thing about math is the different paths to the same answer.

Using log properties,

f(x) = cos ( ln(10) + ln(x) )

Using the cosine addition identity,

f(x) = cos ( ln(10) ) cos( ln(x) ) - sin(ln(10))sin(ln(x))

Note now that anything involving ln(10) is a constant and we can differentiate it as such.

f'(x) = cos ( ln(10) ) (-sin (ln(x))(1/x) - sin(ln(10)) cos(ln(x))/x

f'(x) = (-1/x) cos ( ln(10) ) sin(ln(x)) - (1/x) sin(ln(10))cos(ln(x))

Factor -1/x, we get

f'(x) = (-1/x) [ cos (ln10) sin(ln x) + sin (ln10) cos(ln x) ]

Which is equivalent to just

f'(x) = (-1/x) [ sin (ln10 + ln(x)) ]

f'(x) = (-1/x) [ sin (ln (10x) ) ]

{Yeah, this was the long way, but my mathematical curiosity got the better of me.}

Answer 2

-sin(ln 10x) . (10/10x)

-sin(ln 10x) .(1/x)

-sin(ln 10x) . x^-1

Explanation :

The function of cos[f(x)] is -sin[f(x)] * f'(x)

let ln 10x = f(x)

the derivative of f(x) = f'(x)/f(x)
=10 / 10x
= 1 / x
= x to the power of -1 =x^-1

-sin(ln 10x) . x^-1

Answer 3

let y= f(x)
=> y=cos (ln 10x)
let u= ln 10x
=> also, y= cos u
therefore, dy/du= -sin u

differentiating: u= ln 10x, we have; 1/x
=> du/dx= 1/x

therefore,
dy/dx = dy/du * du/dx
=> dy/dx= (-sin u) * (1/x)
= -sin u/x
but, u= ln 10x
=> dy/dx= -sin (ln 10x) / x

Answer 4

Rewrite as y = 9t^(-a million/2) then you really can prepare the potential rule. Multiply by technique of the exponent (-a million/2), then cut back it by technique of one million from -a million/2 to -3/2. y' = -9/2 t^-3/2 In radical form, this can be y' = -9?t/2t^2