# Use differentials to find an approximate value for the given numbers$y=3\sqrt{1.02}+4\sqrt{1.02}$

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
$3 \sqrt {1.02}+4 \sqrt {1.02}$
Step 1:
Let $y=x^{\large\frac{1}{3}}+x^{\large\frac{1}{4}}$
$dy=\bigg(\large\frac{1}{3}x^{\Large\frac{-2}{3}}+\large\frac{1}{4}x^{\Large\frac{-3}{4}}\bigg)$$dx \quad=\bigg(\large\frac{1}{3x^{\Large\frac{-2}{3}}}+\large\frac{1}{4x^{\Large\frac{-3}{4}}}\bigg)$$dx$
Step 2:
Let $x=1$ and $dx \approx \Delta x=0.02$
Then $f(1)=1^{\large\frac{1}{3}}+1^{\large\frac{1}{4}}$$=2 Also dy=\bigg(\large\frac{1}{3}+\frac{1}{4}\bigg)$$(0.02)$
$\qquad=\large\frac{7}{12}$$(0.02)$
$\qquad=0.0117$
$\therefore f(1.02)=f(1)+dy$
$\qquad=2+0.0117=2.0117$