Q.Find the approximate value of $\tan^{-1}(1.001)$.

Use $f(x+\Delta x)\approx f(x)+f'(x)\Delta x$ with $x=2,\ \Delta x=-0.01$.

$f(x)=x^3-3x+5$, so $f'(x)=3x^2-3$.

Take $x=2$ (nearby whole number) and $\Delta x=1.99-2=-0.01$.

$f(2)=8-6+5=7$ …

$dy$ uses the derivative (linear approximation); $\Delta y$ is the exact change in $y$.

Given $y = e^x + x$, $x=5$, $\Delta x = 0.02$.

Finding $dy$:

$\frac{dy}{dx} = e^x + 1$

$dy = \left(\frac{dy}{dx}\right)_{x=5}\Delta x = (e^5+1)(0.02)$

Using $e^5\approx 148.4132$:

$dy \approx (149.4132)(0.02) \approx 2.9883$

Finding $\Delta y$:

$\Delta y = f(x+\Delta x) - f(x) = \left(e^{5.02}+5.02\right) - \left(e^{5}+5\right) = e^{5.02}-e^{5}+0.02$ …

$\Delta y$ is the exact change in $y$; $dy=f'(x)\Delta x$ is its linear (differential) approximation — the two are close but not identical.

$y=f(x)=x^2+3x+6$, with $x=10$, $\Delta x=0.01$.

Exact change $\Delta y$:

$f(10)=10^2+3(10)+6=100+30+6=136$

$f(10.01)=(10.01)^2+3(10.01)+6=100.2001+30.03+6=136.2301$

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Use the differential/linear approximation $f(a+\Delta x)\approx f(a)+f'(a)\,\Delta x$ with $f(x)=x^{1/3}$ and a nearby perfect cube $a=64$.

$f(x)=x^{1/3},\qquad f(64)=4$

$f'(x) = \frac13 x^{-2/3} \implies f'(64) = \frac13\cdot\frac{1}{16} = \frac{1}{48}$

Take $\Delta x = 65-64 = 1$: …

Use the differential approximation $dV\approx\dfrac{dV}{dx}\,dx$ for a small change $dx$ in the side.

Volume of cube: $V=x^3$, so $\dfrac{dV}{dx}=3x^2$.

Given the side increases by $2\%$: $dx=0.02x$.

$dV\approx 3x^2\cdot(0.02x)=0.06\,x^3$

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For $A=s^2$, a small relative change in the side doubles to give the relative change in area: $\dfrac{dA}{A}\approx 2\dfrac{ds}{s}$.

Let the side of the square be $s$ and area $A=s^2$. Differentiating, $dA = 2s\,ds$.

Relative (percentage) change in area: $\dfrac{dA}{A} = \dfrac{2s\,ds}{s^2} = 2\dfrac{ds}{s}$.

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$\Delta y$ is the exact change in $y$; $dy = y'(x)\,\Delta x$ is its linear (differential) approximation. Compute both at $x=2$, $\Delta x = 0.001$.

Given $y = 5x^2+6x+6$ at $x=2$, $\Delta x = 0.001$.

Exact change $\Delta y$:

$y(2) = 5(4)+6(2)+6 = 20+12+6 = 38$

$y(2.001) = 5(2.001)^2 + 6(2.001) + 6 = 5(4.004001) + 12.006 + 6 = 20.020005+12.006+6 = 38.026005$

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$dy=-\sin x\,\Delta x$ gives the linear (differential) approximation; $\Delta y$ is the exact change, computed from the given data.

Given $y=\cos x$, $x=60^{\circ}$, $\Delta x = 1^{\circ}=0.0174$ radians.

Finding $dy$:

$\frac{dy}{dx} = -\sin x \implies dy = -\sin x\, \Delta x$

At $x=60^{\circ}$, $\sin 60^{\circ} = \dfrac{\sqrt3}{2} \approx 0.8660$:

$dy = -(0.8660)(0.0174) \approx -0.01507$

Finding $\Delta y$:

$\Delta y = f(x+\Delta x) - f(x) = \cos(61^{\circ}) - \cos(60^{\circ})$

Using the given value $\cos61^{\circ}=0.4848$ and $\cos60^{\circ}=0.5$: …

For $A=s^2$, a small relative change in side gives $\frac{dA}{A}=2\frac{ds}{s}$, so a $2\%$ increase in side gives about a $4\%$ increase in area.

Concept: Differentials for approximation

If $A=s^2$, then $dA = 2s\,ds$, so the relative (percentage) change is $\dfrac{dA}{A} = \dfrac{2s\,ds}{s^2} = 2\dfrac{ds}{s}$.

Step 1: Express the given change

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