Find the differential $dy$ and evaluate $dy$ for the given values of $x$ and $dx$\[\]$y=\cos$$x,x=\large\frac{\pi}{6},$$dx=0.05$

Question

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

Answer 1

Toolbox:

• Let $y=f(x)$ be a differentiable function then the quantities $dx$ and $dy$ are called differentials.The differential $dx$ is an independent variable.
• The differential $dy$ is then defined by $dy=f'(x)dx(dx \approx \Delta x)$
• Also $f(x+\Delta x)-f(x)=\Delta y =dy $ from which $f(x+\Delta x) $can be evaluated

Step 1:

$y=\cos x $

$dy= -\sin x dx$

Step 2:

When $x=\large\frac{\pi}{6},$$ dx= 0.05$

$dy=-\sin \large\frac{\pi}{6} $$\times 0.05$

$\quad=\large\frac{-1}{2} $$\times 0.05$

$\quad=-0.025$