Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{f}(\mathrm{x})=\sqrt{25-\mathrm{x}^{2}}$, then what is $\lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(1)}{\mathrm{x}-1}$ equal to
Options:
Solution:
1601 Upvotes
Verified Answer
The correct answer is:
$-\frac{1}{\sqrt{24}}$
$\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}=f^{\prime}(x)$
Now, $f^{\prime}(x)=\frac{-2 x}{2 \sqrt{25-x^{2}}}=\frac{-x}{\sqrt{25-x^{2}}}$
$f^{\prime}(1)=\frac{-1}{\sqrt{25-1}}=\frac{-1}{\sqrt{24}}$
Now, $f^{\prime}(x)=\frac{-2 x}{2 \sqrt{25-x^{2}}}=\frac{-x}{\sqrt{25-x^{2}}}$
$f^{\prime}(1)=\frac{-1}{\sqrt{25-1}}=\frac{-1}{\sqrt{24}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.