Find the domains of the following functions and compute their derivatives:
(\mathbf{a})\ y=\ln(1-x)\qquad \mathrm{(b)~y=ln(4-x^{2})}\qquad\mathrm{(c)~y=ln\left({\frac{x-1}{{x +\ 1}}}\right)-{\frac{1}{4}}x}(a) ln(1 − x) is defined if 1 − x > 0, that is if x < 1. To find its derivative, we use (6.11.2), with h(x) = 1 − x. Then h^{\prime}(x) = −1, and
y=\ln h(x)\Rightarrow y^{\prime}={\frac{h^{\prime}(x)}{h(x)}} (6.11.2)
y^{\prime}={\frac{-1}{1-x}}={\frac{1}{x-1}}(b) ln(4 − x²) is defined if 4 − x² > 0, that is if (2 − x)(2 + x) > 0. This is satisfied if and only if −2 < x < 2. Formula (6.11.2) gives
y^{\prime}={\frac{-2x}{4-x^{2}}}={\frac{2x}{x^{2}-4}}(c) We can write y = ln u − \frac{1}{4}x, where u = (x − 1)/(x + 1). For the function to be defined, we require that u > 0. A sign diagram shows that this is satisfied if x < −1 or x > 1. Using (6.11.2), we obtain
y^{\prime}={\frac{u^{\prime}}{u}}-{\frac{1}{4}}where
u^{\prime}={\frac{1\cdot(x+1)-1\cdot(x-1)}{(x+1)^{2}}}={\frac{2}{(x+1)^{2}}}So
y^{\prime}={\frac{2(x+1)}{(x+1)^{2}(x-1)}}-{\frac{1}{4}}={\frac{9-x^{2}}{4(x^{2}-1)}}={\frac{(3-x)(3+x)}{4(x-1)(x+1)}}