Medium
For some constants $a$ and $b$,find the derivative of $\frac{x-a}{x-b}$.
- A$\frac{a-b}{(x-b)^{2}}$
- B$\frac{b-a}{(x-b)^{2}}$
- C$\frac{a+b}{(x-b)^{2}}$
- D$\frac{a-b}{(x-a)^{2}}$
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Similar Questions
For the function $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$,prove that $f^{\prime}(1) = 100 f^{\prime}(0)$.
Medium
View SolutionIf $y = (1 + x^2)\tan^{-1}x - x,$ then $\frac{dy}{dx} = $
Easy
View SolutionMatch the functions in List-$I$ with their derivatives given in List-$II$.
List-$I$ List-$II$ $A$. $\sec^{-1} x$ $I$. $\frac{1}{1-x^2}, x \in (-1, 1)$ $B$. $\tanh^{-1} x$ $II$. $\frac{-1}{|x| \sqrt{x^2+1}}, x \neq 0$ $C$. $\coth^{-1} x$ $III$. $\frac{1}{|x| \sqrt{x^2-1}}, |x| > 1$ $D$. $\operatorname{cosech}^{-1} x$ $IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$ $V$. $\frac{-1}{|x| \sqrt{1-x^2}}, |x| < 1, x \neq 0$
| List-$I$ | List-$II$ |
| $A$. $\sec^{-1} x$ | $I$. $\frac{1}{1-x^2}, x \in (-1, 1)$ |
| $B$. $\tanh^{-1} x$ | $II$. $\frac{-1}{|x| \sqrt{x^2+1}}, x \neq 0$ |
| $C$. $\coth^{-1} x$ | $III$. $\frac{1}{|x| \sqrt{x^2-1}}, |x| > 1$ |
| $D$. $\operatorname{cosech}^{-1} x$ | $IV$. $\frac{1}{1-x^2}, x \in R - [-1, 1]$ |
| $V$. $\frac{-1}{|x| \sqrt{1-x^2}}, |x| < 1, x \neq 0$ |
$\frac{d}{d x} \left\{ (1+x^2) \tan^{-1}(x) \right\} =$
The derivative of ${x^6} + {6^x}$ with respect to $x$ is:
Easy
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