Browse Questions

# Find the differential $dy$ and evaluate $dy$ for the given values of $x$ and $dx$$y=(x^{2}+5)^{3},x=1,dx=0.1$

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

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:
$y=(x^2+5)^3$
$dy=3(x^2+5)^2-2xdx$
$\quad=6x(x^2+5)^2dx$
Step 2:
When $x=1,dx=0.05$
$dy= 6(6)^2(0.05)$
$\quad=216 (0.05)$
$\quad=10.8$