How do you differentiate $ln ∣ ∣ \frac{cos x}{cos x} - 1 ∣ ∣$ $ln |cosx/cosx-1|$ ?

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Answer 1 · 2017-07-09T23:28:56

$\frac{d}{d x} [ln ∣ ∣ \frac{cos x}{cos x} - 1 ∣ ∣] = 0$ $d/(dx) [ln|(cosx)/(cosx) - 1|] = color(blue)(0$

We're asked to find the derivative

$\frac{d}{d x} [ln ∣ ∣ \frac{cos x}{cos x} - 1 ∣ ∣]$ $d/(dx) [ln|(cosx)/(cosx) - 1|]$

The value $\frac{cos x}{cos x} = 1$ $(cosx)/(cosx) = 1$ :

$\frac{d}{d x} [ln | 1 - 1 |$ $d/(dx)[ln|1-1|$

$\frac{d}{d x} [ln | 0 |]$ $d/(dx)[ln|0|]$

The natural logarithm of $0$ $0$ is $- \infty$ $-oo$ :

$\frac{d}{d x} [- \infty]$ $d/(dx)[-oo]$

The derivative of negative infinity (a constant) is $0$ $0$ :

$\frac{d}{d x} [ln ∣ ∣ \frac{cos x}{cos x} - 1 ∣ ∣] = 0$ $d/(dx) [ln|(cosx)/(cosx) - 1|] = color(blue)(0$

How do you differentiate ln∣∣cosxcosx−1∣∣ln |cosx/cosx-1| ?