Application of Derivatives

$\text{ We need to differentiate }\cos \left(\sqrt{x}\right)\text{ with respect to }x\text{. }$

Step 1: Apply Chain Rule

Let: $y=\cos \left(\sqrt{x}\right)$

$$\begin{gathered} \text{ Define }u=\sqrt{x}\Rightarrow x^{1/2}\text{, so that: } \\ y=\cos \left(u\right) \end{gathered}$$

Now, differentiate both sides:

$\frac{dy}{dx}=\frac{d}{du}\cos \left(u\right)\cdot \frac{du}{dx}$

Step 2: Compute the Derivatives

$$\begin{gathered} \frac{d\left(\cos u\right)}{du}=-\sin \left(u\right) \\ \frac{du}{dx}=\frac{d \left(x^{\left({\displaystyle \frac{1}{2}}\right)}\right)}{dx} \end{gathered}$$


Step 3: Multiply the Terms

$\frac{dy}{dx}=-\frac{\sin \sqrt{x}}{2\sqrt{x}}$

Given: $f\left(x\right)=x^3-3x^2+12x-18$

Then, $f^{\text{'}}\left(x\right)=3x^2-6x+12=3\left(x^2-2x+4\right)$

since, $x^2-2x+4=(x-1{)}^2+3>0 \forall x\in \mathrm{ℝ}$

$\Rightarrow f^{\text{'}}\left(x\right)>0 \forall x\in \mathrm{ℝ}$

\(\Rightarrow f\) is strictly increasing on \(\mathbb{R}\)

$r=10 \mathrm{cm}, \frac{dV}{dt}=100\pi { \mathrm{cm}}^3/\mathrm{s}$

$V=\pi r^{2}h=\pi (10{)}^{2}h=100\pi h$

$\frac{dV}{dt}=100\pi \frac{dh}{dt}$

$\Rightarrow \frac{dh}{dt}=\frac{100\pi }{100\pi }=1 \mathrm{cm}/\mathrm{s}$

$f\left(x\right)=\frac{x}{2}+\frac{2}{x}$

$f^{\text{'}}\left(x\right)=\frac{d}{dx}\left(\frac{1}{2}x+2x^{-1}\right)=\frac{1}{2}-2x^{-2}=\frac{1}{2}-\frac{2}{x^2}$

Critical points occur where the first derivative is equal to zero.

$$\begin{gathered} \frac{1}{2}-\frac{2}{x^2}=0 \\ \frac{1}{2}-\frac{2}{x^2}=0 \\ {x}^2=4 \\ x=\pm 2 \end{gathered}$$

Find the second derivative.

$f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left(\frac{1}{2}-2x^{-2}\right)=0-2(-2)x^{-3}=\frac{4}{x^3}$

$f^{\text{'}\text{'}}\left(2\right)=\frac{4}{2^3}=\frac{4}{8}=\frac{1}{2}$, since $f^{\text{'}\text{'}}\left(2\right)>0$ there is a local minima at $x=2$.

$f^{\text{'}\text{'}}\left(-2\right)=\frac{4}{{\left(-2\right)}^3}=\frac{4}{-8}=-\frac{1}{2}$, since $f^{\text{'}\text{'}}(-2)<0$,there is a local maxima at $x=-2$.

Given, $r=10 \mathrm{cm}, \frac{dV}{dt}=100\pi { \mathrm{cm}}^3/\mathrm{s}$

We know, $V=\pi r^{2}h=\pi (10{)}^{2}h=100\pi h$

$\frac{dV}{dt}=100\pi \frac{dh}{dt}$

$\Rightarrow \frac{dh}{dt}=\frac{100\pi }{100\pi }=1 \mathrm{cm}/\mathrm{s}$

Given: $f\left(x\right)=\frac{x}{2}+\frac{2}{x}$

Then, $f^{\text{'}}\left(x\right)=\frac{d}{dx}\left(\frac{1}{2}x+2x^{-1}\right)=\frac{1}{2}-2x^{-2}=\frac{1}{2}-\frac{2}{x^2}$

Critical points occur where the first derivative is equal to zero.

$$\begin{gathered} \frac{1}{2}-\frac{2}{x^2}=0 \\ \Rightarrow x^2=4 \\ \Rightarrow x=\pm 2 \end{gathered}$$

Now,

$f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left(\frac{1}{2}-2x^{-2}\right)=0-2\left(-2\right)x^{-3}=\frac{4}{x^3}$

$f^{\text{'}\text{'}}\left(2\right)=\frac{4}{2^3}=\frac{4}{8}=\frac{1}{2}$, since $f^{\text{'}\text{'}}\left(2\right)>0$ there is a local minima at $x=2$.

$f^{\text{'}\text{'}}\left(-2\right)=\frac{4}{{\left(-2\right)}^3}=\frac{4}{-8}=-\frac{1}{2}$, since $f^{\text{'}\text{'}}\left(-2\right)<0$,there is a local maxima at $x=-2$.

$\frac{dy}{dx}=\frac{d}{dx}\left(7x-x^3\right)=7-3x^2$

Let $m=\frac{dy}{dx}$.

$\frac{dm}{dt}=\frac{d}{dt}\left(7-3x^2\right)=-6x\frac{dx}{dt}$

Substitute $x=5$ and $\frac{dx}{dt}=2$.

$\frac{dm}{dt}=-6\left(5\right)\left(2\right)=-60$

Given, $f\left(x\right)=x^{2025}-x^{2000}$

$f\text{'}\left(x\right)=2025x^{20224}-2020x^{1999}$

for critical points :

$f\text{'}\left(x\right)=0$

$\Rightarrow x={\left(\frac{2000}{2025}\right)}^{1/25}={\left(\frac{80}{81}\right)}^{1/25}=\alpha$ (say)

Now,

$f\left(x\right)=x^{2000}\left(x^{25}-1\right)$

$\therefore f\left(0\right)=0,f\left(1\right)=0,$

$f\left(\alpha \right)={\left(\frac{80}{81}\right)}^{80}\cdot \frac{-1}{81}={80}^{80}\cdot {(-81)}^{-81}$

Hence, \(n=-81\)

$f\left(x\right)=x^2-4x+6$

$f^{\text{'}}\left(x\right)=2x-4$

$f^{\text{'}}\left(x\right)>0\Rightarrow 2x-4>0\Rightarrow x>2$

The interval of increasing is $(2,\infty )$.

$f\left(x\right)=x^3-3x^2+12x-18$

$f^{\text{'}}\left(x\right)=3x^2-6x+12=3\left(x^2-2x+4\right)$

$x^2-2x+4=(x-1{)}^2+3>0 \forall x\in \mathrm{ℝ}$

$\Rightarrow f^{\text{'}}\left(x\right)>0 \forall x\in \mathrm{ℝ}$

\(\Rightarrow f\) is strictly increasing on \(\mathbb{R}\)

A function \(f(x)\) is strictly increasing on an interval \((a, b)\) if for any two points \(x_1, x_2 \in(a, b)\), with \(x_1<x_2\), we have \(f\left(x_1\right)<f\left(x_2\right)\). This means that the derivative of the function, \(f^{\prime}(x)\), must be positive for all \(x\) in ( \(a, b\) ). Therefore, the correct condition for \(f(x)\) to be strictly increasing in \((a, b)\) is \(f^{\prime}(x)>0\) for all \(x\) in \((a, b)\).

Given: $\mathrm{y}=7x-x^3$

Let $m=\frac{dy}{dx}$

$=\frac{d}{dx}\left(7x-x^3\right)=7-3x^2$

Now, rate of change of slope,

$\frac{dm}{dt}=\frac{d}{dt}\left(7-3x^2\right)=-6x\frac{dx}{dt}$

Substitute $x=5$ and $\frac{dx}{dt}=2$.

$\frac{dm}{dt}=-6\left(5\right)\left(2\right)=-60$ units per second

Let \(x_1<x_2 \in(a, b)\).
By MVT, there exists \(\ c\) in \(\left(x_1, x_2\right)\) with \(\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=f^{\prime}(c)\).
Assume \(f^{\prime}(x)>0 \ \forall x \in(a, b)\).
Then \(f^{\prime}(c)>0\) and \(x_2-x_1>0\).
\(\Rightarrow f\left(x_2\right)-f\left(x_1\right)>0\).
\(\Rightarrow f\left(x_2\right)>f\left(x_1\right)\) (strictly increasing).

$f\left(x\right)=x^2-4x+6$

$f^{\text{'}}\left(x\right)=2x-4$

$f^{\text{'}}\left(x\right)>0\Rightarrow 2x-4>0\Rightarrow x>2$

The interval of increase is $(2,\infty )$

$\text{ We need to differentiate }\cos \left(\sqrt{x}\right)\text{ with respect to }x\text{. }$

Step 1: Apply Chain Rule

Let: $y=\cos \left(\sqrt{x}\right)$

$$\begin{gathered} \text{ Define }u=\sqrt{x}\Rightarrow x^{1/2}\text{, so that: } \\ y=\cos \left(u\right) \end{gathered}$$

Now, differentiate both sides:

$\frac{dy}{dx}=\frac{d}{du}\cos \left(u\right)\cdot \frac{du}{dx}$

Step 2: Compute the Derivatives

$$\begin{gathered} \frac{d\left(\cos u\right)}{du}=-\sin \left(u\right) \\ \frac{du}{dx}=\frac{d \left(x^{\left({\displaystyle \frac{1}{2}}\right)}\right)}{dx} \end{gathered}$$


Step 3: Multiply the Terms

$\frac{dy}{dx}=-\frac{\sin \sqrt{x}}{2\sqrt{x}}$