Using differentials,find the approximate valu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $y = x^{\frac{1}{4}}$. Let $x = \frac{16}{81}$ and $\Delta x = \frac{1}{81}$.Then,$\Delta y = (x + \Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} =

Vedclass · Accepted Answer

$0.677$

Vedclass · Answer

(A) Let $y = x^{\frac{1}{4}}$. Let $x = \frac{16}{81}$ and $\Delta x = \frac{1}{81}$.Then,$\Delta y = (x + \Delta x)^{\frac{1}{4}} - x^{\frac{1}{4}} = \left(\frac{17}{81}ight)^{\frac{1}{4}} - \left(\frac{16}{81}ight)^{\frac{1}{4}} = \left(\frac{17}{81}ight)^{\frac{1}{4}} - \frac{2}{3}$.Thus,$\left(\frac{17}{81}ight)^{\frac{1}{4}} = \frac{2}{3} + \Delta y$.Using the differential $dy \approx \Delta y$,we have $dy = \frac{dy}{dx} \Delta x = \frac{1}{4} x^{-\frac{3}{4}} \Delta x$.Substituting $x = \frac{16}{81}$ and $\Delta x = \frac{1}{81}$:$dy = \frac{1}{4 \left(\frac{16}{81}ight)^{\frac{3}{4}}} 	imes \frac{1}{81} = \frac{1}{4 	imes \frac{8}{27}} 	imes \frac{1}{81} = \frac{27}{32} 	imes \frac{1}{81} = \frac{1}{32 	imes 3} = \frac{1}{96} \approx 0.0104$.Therefore,the approximate value is $\frac{2}{3} + 0.0104 = 0.6667 + 0.0104 = 0.6771 \approx 0.677$.

Using differentials,find the approximate value of $\left(\frac{17}{81}\right)^{\frac{1}{4}}$.

Similar Questions

The approximate value of the function $f(x) = x^{3} + 5x^{2} - 7x + 10$ at $x = 1.1$ is (in $.6$)

Using differentials,find the approximate value of $\sqrt{0.082}$. Let $f(x) = \sqrt{x}$.

Find the approximate value of $f(5.001),$ where $f(x)=x^{3}-7x^{2}+15$.

The approximate value of $\sqrt[3]{64.04}$ is

The approximate value of $\tan ^{-1}(0.999)$ (up to $4$ decimal places) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module