Q: \$\displaystyle \frac{d(\tan x.)}{dx}\$

\$\displaystyle y=\tan x.\$\$\displaystyle \therefore \dfrac{dy}{dx} = \lim_{\delta x\rightarrow 0}\dfrac{\tan \left ( x+\delta x \right )-\tan x}{\delta x}\$(Change to \$\sin\$ and \$\cos\$).\$\displaystyle =\lim_{\delta x\rightarrow 0}\dfrac{\sin \left ( x+\delta x-x \right )}{\delta x.\cos x\cos \left(x+\delta x \right)}\$\$\displaystyle =\lim_{\delta x\rightarrow 0}\dfrac{\sin \delta x}{\delta x}.\dfrac{1}{\cos x\cos \left(x+\delta x \right)}\$\$\displaystyle =1. \sec^{2}x= \sec^{2}x.\$

Q: \$\displaystyle \frac{d(\sin x)}{dx}.\$

\$\displaystyle y= \sin x= f\left ( x \right ),\$(say)\$\displaystyle y= \delta y = \sin \left ( x+\delta x \right )= f\left ( x+\delta x \right )\$\$\displaystyle \therefore \delta y = \left ( x+\delta x \right )-\sin x= f\left ( x+\delta x \right )-f (x)\$\$\displaystyle \therefore \frac{\delta y}{\delta x}= \frac{\sin \left ( x+\delta x \right )-\sin x}{\delta x}= \frac{f\left ( x+\delta x \right )-f\left ( x \right )}{\delta x}\$\$\displaystyle \therefore \frac{\delta y}{\delta x}= \lim_{\delta x\rightarrow 0} \frac{\sin \left ( x+\delta x \right )-\sin x}{\delta x}\$\$\displaystyle = lim_{\delta x\rightarrow 0} = \frac{f\left ( x+\delta x \right )-f\left ( x \right )}{\delta x}\$\$\displaystyle = \lim_{\delta x\rightarrow 0} = \frac{2\cos \left ( x+\frac{1}{2}\delta x \right )\sin\frac{1}{2}\delta x}{\delta x}\$\$\displaystyle = \lim_{\delta x\rightarrow 0}\cos \left ( x+\frac{\delta x}{2}\right )\frac{\sin \left ( \delta x/2 \right )}{\left (\delta x/2 \right )}\$\$\displaystyle =\cos x.1= \cos x \left ( \because \lim_{\delta x\rightarrow 0} \frac{\sin \theta }{\theta }= 1.\right )\$

Q: Differentiate\$\displaystyle \cos x\$

\$y=\cos x\$\$\displaystyle\frac{dy}{dx}=\lim _{ h\rightarrow 0 }{ \frac { f(x+h)-f(x) }{ h } } \$\$=\displaystyle\lim _{ h\rightarrow 0 }{ \frac {\cos(x+h)-\cos(x)}{ h } } \$\$=\displaystyle\lim _{ h\rightarrow 0 }{ \frac {-2\sin\left(\displaystyle\frac{2x+h}{2}\right)\sin(h/2)}{ h } } \$\$=-\sin x\$\$\therefore \displaystyle\frac{dy}{dx}=-\sin x\$

Q: Say true or false.If \$y=2 \sec x\$, then \$\dfrac{dy}{dx} \$ is \$2 \sec x \tan x\$.

If \$y=2 \sec x\$Then, the derivative of \$y\$ with respect to \$x\$ is \$\dfrac{dy}{dx}=2 \sec x \tan x\$.Therefore, the given statement is true.

Q: \$\displaystyle \dfrac{d}{dx} \sec x=\$

\$\displaystyle y= \sec x\$\$\displaystyle \therefore \frac{dy}{dx}=\lim_{\delta x\rightarrow 0}\frac{\sec \left ( x+\delta x \right )-\sec x}{\delta x}\$ ....(Change to cos)\$\displaystyle = \lim_{\delta x\rightarrow 0}\frac{\cos x- \cos \left ( x+\delta x \right )}{\delta x.\cos\left ( x+\delta x \right )\cos x}\$ We have used \$\cos C-\cos D\$, formula\$\displaystyle = \sin x.\frac{1}{\cos x \cos x}= \sec x\tan x\$

Q: State if the given statement is True or FalseDerivative of \$y= \cos x\$ with respect to \$x\$ is \$\sin x\$.

If \$y=cos x\$Then, the derivative of \$y\$ with respect to \$x\$ i.e., \$\dfrac{dy}{dx}=-\sin x\$.Therefore, the given statement is false.

Q: \$\displaystyle \frac{d\sin x^{2}}{dx}\$

\$\displaystyle y= \sin x^{2}\$\$\displaystyle \frac{dy}{dx}= \lim_{\delta x\rightarrow 0}\frac{\sin \left ( x+\delta x \right )^{2}-\sin x^2}{\delta x}\$\$\displaystyle =\lim_{\delta x\rightarrow 0}\frac{\sin \left ( x^{2}+2x\delta x^{2}+(\delta x)^2\right )-\sin x^{2}}{\delta x}\$\$=\displaystyle \lim_{\delta x\rightarrow 0}\frac{2\cos \left ( x^{2}+x\delta x+\frac{1}{2}\delta x^{2} \right )\sin \left ( x\delta x+\frac{1}{2}\delta x^{2} \right )}{\delta x}\$\$=\displaystyle 2\cos x^{2}\lim_{\delta x\rightarrow 0}\frac{\sin \left ( x\delta x+\frac{1}{2}\delta x^{2} \right )}{x\delta x+\frac{1}{2}\delta x^{2}}.\frac{x\delta x+\frac{1}{2}\delta x^{2}}{\delta x}\$\$\displaystyle =2\cos x^{2}.1.x = 2x\cos x^{2}\$

Q: \$\displaystyle \frac{d}{dx}(\sin^{2}x)\$

\$\displaystyle y= \sin ^{2}x\$\$\displaystyle \frac{dy}{dx}=\lim_{\delta x\rightarrow 0}\frac{\sin ^{2}\left ( x+\delta x \right )-\sin ^{2}x}{\delta x}\$\$\displaystyle =\lim_{\delta x\rightarrow 0}\frac{\sin ^{2}\left ( x+\delta x +x\right ).\sin \left ( x+\delta x-x \right )}{\delta x}\$\$\displaystyle =\lim_{\delta x\rightarrow 0}\sin \left ( 2x+\delta x \right )\frac{\sin \delta x}{\delta x}= \sin 2x.\$We have used\$[\displaystyle \sin^2 A-\sin ^{2}B= \sin \left ( A+B \right )\sin \left ( A-B \right )]\$

Q: \$\displaystyle \lim_{x\rightarrow 0}x^{2}\displaystyle \sin\frac{\pi}{x}=\$

\$sin\ \frac{\pi }{x} \text{ behaves\ as\ oscillatory\ function\ at} \$\$x\rightarrow 0\$\$but\ have\ finite\ value = 0\$\$0\times\ finite\ value\ =0\$

Q: If \$y=2 \sin x -3x^4 + 8\$, then \$\dfrac{dy}{dx}\$ is

If \$y=2 \sin x -3x^4 + 8\$\$\Rightarrow \cfrac{dy}{dx}=2 \cos x-3(4x^{(4-1)})+0\$\$\Rightarrow \cfrac{dy}{dx}=2 \cos x-12x^3\$Therefore, the correct answer is \$B\$.

Question 1

$\displaystyle \frac{d(\tan x.)}{dx}$

Accepted Answer

$\displaystyle y=\tan x.$$\displaystyle \therefore \dfrac{dy}{dx} = \lim_{\delta x\rightarrow 0}\dfrac{\tan \left ( x+\delta x \right )-\tan x}{\delta x}$(Change to $\sin$ and $\cos$).$\displaystyle =\lim_{\delta x\rightarrow 0}\dfrac{\sin \left ( x+\delta x-x \right )}{\delta x.\cos x\cos \left(x+\delta x \right)}$$\displaystyle =\lim_{\delta x\rightarrow 0}\dfrac{\sin \delta x}{\delta x}.\dfrac{1}{\cos x\cos \left(x+\delta x \right)}$$\displaystyle =1. \sec^{2}x= \sec^{2}x.$

Question 2

$\displaystyle \frac{d(\sin x)}{dx}.$

Accepted Answer

$\displaystyle y= \sin x= f\left ( x \right ),$(say)$\displaystyle y= \delta y = \sin \left ( x+\delta x \right )= f\left ( x+\delta x \right )$$\displaystyle \therefore \delta y =  \left ( x+\delta x \right )-\sin x= f\left ( x+\delta x \right )-f (x)$$\displaystyle \therefore \frac{\delta y}{\delta x}= \frac{\sin \left ( x+\delta x \right )-\sin x}{\delta x}= \frac{f\left ( x+\delta x \right )-f\left ( x \right )}{\delta x}$$\displaystyle \therefore \frac{\delta y}{\delta x}= \lim_{\delta x\rightarrow 0} \frac{\sin \left ( x+\delta x \right )-\sin x}{\delta x}$$\displaystyle  = lim_{\delta x\rightarrow 0} = \frac{f\left ( x+\delta x \right )-f\left ( x \right )}{\delta x}$$\displaystyle = \lim_{\delta x\rightarrow 0} = \frac{2\cos \left ( x+\frac{1}{2}\delta x \right )\sin\frac{1}{2}\delta x}{\delta x}$$\displaystyle = \lim_{\delta x\rightarrow 0}\cos \left ( x+\frac{\delta x}{2}\right )\frac{\sin \left ( \delta x/2 \right )}{\left (\delta x/2  \right )}$$\displaystyle =\cos x.1= \cos x \left ( \because \lim_{\delta x\rightarrow 0} \frac{\sin \theta }{\theta }= 1.\right )$

Question 3

Differentiate$\displaystyle \cos x$

Accepted Answer

$y=\cos x$$\displaystyle\frac{dy}{dx}=\lim _{ h\rightarrow 0 }{ \frac { f(x+h)-f(x) }{ h }  } $$=\displaystyle\lim _{ h\rightarrow 0 }{ \frac {\cos(x+h)-\cos(x)}{ h }  } $$=\displaystyle\lim _{ h\rightarrow 0 }{ \frac {-2\sin\left(\displaystyle\frac{2x+h}{2}\right)\sin(h/2)}{ h }  } $$=-\sin x$$\therefore \displaystyle\frac{dy}{dx}=-\sin x$

Question 4

Say true or false.If $y=2  \sec  x$, then $\dfrac{dy}{dx} $ is $2 \sec x  \tan x$.

Accepted Answer

If $y=2 \sec  x$Then, the derivative of $y$ with respect to $x$ is $\dfrac{dy}{dx}=2 \sec x  \tan x$.Therefore, the given statement is true.

Question 5

$\displaystyle \dfrac{d}{dx} \sec x=$

Accepted Answer

$\displaystyle y= \sec x$$\displaystyle \therefore \frac{dy}{dx}=\lim_{\delta x\rightarrow 0}\frac{\sec \left ( x+\delta x \right )-\sec x}{\delta x}$          ....(Change to cos)$\displaystyle = \lim_{\delta x\rightarrow 0}\frac{\cos x- \cos \left ( x+\delta x \right )}{\delta x.\cos\left ( x+\delta x \right )\cos x}$ We have used $\cos C-\cos D$, formula$\displaystyle = \sin x.\frac{1}{\cos x \cos x}= \sec x\tan x$

Question 6

State if the given statement is True or FalseDerivative of $y= \cos  x$ with respect to $x$ is $\sin x$.

Accepted Answer

If $y=cos x$Then, the derivative of $y$ with respect to $x$ i.e., $\dfrac{dy}{dx}=-\sin x$.Therefore, the given statement is false.

Question 7

$\displaystyle \frac{d\sin x^{2}}{dx}$

Accepted Answer

$\displaystyle y= \sin x^{2}$$\displaystyle \frac{dy}{dx}= \lim_{\delta x\rightarrow 0}\frac{\sin \left ( x+\delta x \right )^{2}-\sin x^2}{\delta x}$$\displaystyle =\lim_{\delta x\rightarrow 0}\frac{\sin \left ( x^{2}+2x\delta x^{2}+(\delta x)^2\right )-\sin x^{2}}{\delta x}$$=\displaystyle \lim_{\delta x\rightarrow 0}\frac{2\cos \left ( x^{2}+x\delta x+\frac{1}{2}\delta x^{2} \right )\sin \left ( x\delta x+\frac{1}{2}\delta x^{2} \right )}{\delta x}$$=\displaystyle 2\cos x^{2}\lim_{\delta x\rightarrow 0}\frac{\sin \left ( x\delta x+\frac{1}{2}\delta x^{2} \right )}{x\delta x+\frac{1}{2}\delta x^{2}}.\frac{x\delta x+\frac{1}{2}\delta x^{2}}{\delta x}$$\displaystyle =2\cos x^{2}.1.x = 2x\cos x^{2}$

Question 8

$\displaystyle \frac{d}{dx}(\sin^{2}x)$

Accepted Answer

$\displaystyle y= \sin ^{2}x$$\displaystyle \frac{dy}{dx}=\lim_{\delta x\rightarrow 0}\frac{\sin ^{2}\left ( x+\delta x \right )-\sin ^{2}x}{\delta x}$$\displaystyle =\lim_{\delta x\rightarrow 0}\frac{\sin ^{2}\left ( x+\delta x +x\right ).\sin \left ( x+\delta x-x \right )}{\delta x}$$\displaystyle =\lim_{\delta x\rightarrow 0}\sin \left ( 2x+\delta x \right )\frac{\sin \delta x}{\delta x}= \sin 2x.$We have used$[\displaystyle \sin^2 A-\sin ^{2}B= \sin \left ( A+B \right )\sin \left ( A-B \right )]$

Question 9

$\displaystyle \lim_{x\rightarrow 0}x^{2}\displaystyle \sin\frac{\pi}{x}=$

Accepted Answer

$sin\ \frac{\pi }{x} \text{ behaves\ as\ oscillatory\ function\ at} $$x\rightarrow 0$$but\ have\ finite\ value = 0$$0\times\ finite\ value\ =0$

Question 10

If $y=2  \sin  x -3x^4 + 8$, then $\dfrac{dy}{dx}$ is

Accepted Answer

If $y=2  \sin  x -3x^4 + 8$$\Rightarrow \cfrac{dy}{dx}=2  \cos  x-3(4x^{(4-1)})+0$$\Rightarrow \cfrac{dy}{dx}=2  \cos  x-12x^3$Therefore, the correct answer is $B$.

Limits And Derivatives

Maths
5010 Views

$\frac{d ( tan x . )}{d x}$

$\frac{d ( sin x )}{d x} .$

Differentiate
$cos x$

Say true or false.
If $y = 2 sec x$ , then $\frac{d y}{d x}$ is $2 sec x tan x$ .

$\frac{d}{d x} sec x =$

State if the given statement is True or False
Derivative of $y = cos x$ with respect to $x$ is $sin x$ .

$\frac{d sin x ^{2}}{d x}$

$\frac{d}{d x} (sin^{2} x)$

$x \to 0 lim x^{2} sin \frac{π}{x} =$

If $y = 2 sin x - 3 x^{4} + 8$ , then $\frac{d y}{d x}$ is

Limits And Derivatives

Maths 5010 Views

dxd(tanx.)​

dxd(sinx)​.

Differentiate cosx

Say true or false.If y=2secx, then dxdy​ is 2secxtanx.

dxd​secx=

State if the given statement is True or FalseDerivative of y=cosx with respect to x is sinx.

dxdsinx2​

dxd​(sin2x)

x→0lim​x2sinxπ​=

If y=2sinx−3x4+8, then dxdy​ is