$\frac{a+b \sin x}{c+d \cos x}$
[question] Question. $\frac{a+b \sin x}{c+d \cos x}$ [/question] [solution] solution: Given $y=\frac{a+b \sin x}{c+d \cos x}$ Applying division rule or quotient rule of differentiation that is $\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{t}}{\mathrm{y}}\right)=\frac{\mathrm{y} \cdot \mathrm{dt}}{\mathrm{dx}}-\mathrm{t} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$ $\Rightarrow y=\frac{a+b \sin x}{c+d \cos x}$ $\Rightarrow \frac{d y}{d x}=\frac{(c+d \cos x) \frac{d}{d x}(a+b \sin x)-(a...
$\frac{x^{5}-\cos x}{\sin x}$
[question] Question. $\frac{x^{5}-\cos x}{\sin x}$ [/question] [solution] solution: Given $y=\frac{x^{5}-\cos x}{\sin x}$ $\mathrm{d} / \mathrm{dx}\left(\mathrm{x}^{5}-\cos \mathrm{x}\right) / \sin \mathrm{x}=\left[\sin \mathrm{x} \cdot \mathrm{d} / \mathrm{dx}\left(\mathrm{x}^{5}-\cos \right.\right.$ $\left.x)-\left(x^{5}-\cos x\right) \cdot d / d x(\sin x)\right] / \sin ^{2} x$ By using quotient rule, $=\left[\sin x\left(5 x^{4}+\sin x\right)-\right.$ $\left.\left(x^{5}-\cos x\right)(\cos x)\r...
$\frac{3 x+4}{5 x^{2}-7 x+9}$
[question] Question. $\frac{3 x+4}{5 x^{2}-7 x+9}$ [/question] [solution] solution: Given $\mathrm{y}=\frac{3 \mathrm{x}+4}{5 \mathrm{x}^{2}-7 \mathrm{x}+9}$ Applying quotient rule of differentiation that is $\Rightarrow \frac{d}{d x}\left(\frac{t}{y}\right)=\frac{y \cdot \frac{d t}{d x}-t \cdot \frac{d y}{d x}}{y^{2}}$ Applying the rule $\Rightarrow \frac{d y}{d x}=\frac{\left(5 x^{2}-7 x+9\right) \frac{d}{d x}(3 x+4)-(3 x+4) \frac{d}{d x}\left(5 x^{2}-7 x+9\right)}{\left(5 x^{2}-7 x+9\right)^{...
$x+\frac{1}{x}^{3}$
[question] Question. $x+\frac{1}{x}^{3}$ [/question] [solution] solution: Let $y=\left(x+\frac{1}{x}\right)^{3}$ Now differentiating y with respect to $x$ we get $\Rightarrow$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{3}$ Expanding the equation using $(a+b)^{3}$ formula then we get $=\frac{d}{d x}\left(x^{3}+\frac{1}{x^{3}}+3 x+\frac{3}{x}\right)$ Splitting the differential we get $=\frac{d}{d x}\left(x^{3}\right)+\frac{d}{d x}\le...
$\frac{x^{4}+x^{3}+x^{2}+1}{x}$
[question] Question. $\frac{x^{4}+x^{3}+x^{2}+1}{x}$ [/question] [solution] solution: Let $y=\frac{x^{4}+x^{3}+x^{2}+1}{x}$ $\Rightarrow$$y=\frac{x^{4}+x^{3}+x^{2}+1}{x}$ Dividing by $x$ we get $\Rightarrow$$y=x^{3}+x^{2}+x+\frac{1}{x}$ Differentiating given equation with respect to $x$ $\Rightarrow$$\frac{d y}{d x}=\frac{d}{d x}\left(x^{3}+x^{2}+x+\frac{1}{x}\right)$ On differentiation we get $\Rightarrow$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{3}+\mathrm{x}^{2}+\mathrm{x}+\frac{1}{\ma...
$\lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}$
[question] Question. $\lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}$ [/question] [solution] solution: Given $\lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}$ Now by splitting the limits in above equation we get $\Rightarrow$$\lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}=\lim _{x \rightarrow 0} \frac{\sin x}{x}-\lim _{x \rightarrow 0} \frac{2 \sin 3 x}{x}+\lim _{x \rightarrow 0} \frac{\sin 5 x}{x}$ Taking constant term outside the limits we get $\Rightar...
$\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x}$
[question] Question. $\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x}$ [/question] [solution] solution: Given $\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x}$ Multiply and divide the given equation by $\sqrt{2}-\sqrt{1+\cos x}$ Then we get $\Rightarrow$ $\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x}=\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x} \times\left(\frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\co...
$\lim _{x \rightarrow \frac{\pi}{6}} \frac{\cot ^{2} x-3}{\operatorname{cosec} x-2}$
[question] Question. $\lim _{x \rightarrow \frac{\pi}{6}} \frac{\cot ^{2} x-3}{\operatorname{cosec} x-2}$ [/question] [solution] solution: Given $\lim _{x \rightarrow \frac{\pi}{6}} \frac{\cot ^{2} x-3}{\operatorname{cosec} x-2}$ We know that $\cot ^{2} x=\operatorname{cosec}^{2} x-1$ By using this in given equation we get $\Rightarrow$ $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\left(\operatorname{cosec}^{2} x-1\right)-3}{\operatorname{cosec} x-2}=\lim _{x \rightarrow \frac{\pi}{4}} \frac{\oper...
$\lim _{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}}$
[question] Question. $\lim _{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}}$ [/question] [solution] solution: Given $\lim _{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}}$ Now we have to rationalize the denominator by multiplying the dividing by its rationalizing factor then we get $\Rightarrow$$\lim _{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}}=\lim _{x \rightarrow a}\left[\frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}} \times \frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}...
$\lim _{x \rightarrow 0} \frac{\sin 2 x+3 x}{2 x+\tan 3 x}$
[question] Question. $\lim _{x \rightarrow 0} \frac{\sin 2 x+3 x}{2 x+\tan 3 x}$ [/question] [solution] solution: Given $\lim _{x \rightarrow 0} \frac{\sin 2 x+3 x}{2 x+\tan 3 x}$ Multiply and divide the numerator of given equation by $2 x$ $\Rightarrow$$\lim _{x \rightarrow 0} \frac{\sin 2 x+3 x}{2 x+\tan 3 x}=\lim _{x \rightarrow 0} \frac{2 x(\sin 2 x) / 2 x+3 x}{2 x+3 x(\tan 3 x) / 3 x}$ Now by splitting the limits we get $\Rightarrow$$\lim _{x \rightarrow 0} \frac{2 x(\sin 2 x) / 2 x+3 x}{2 ...
$\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sqrt{3} \sin x-\cos x}{x-\frac{\pi}{6}}$
[question] Question. $\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sqrt{3} \sin x-\cos x}{x-\frac{\pi}{6}}$ [/question] [solution] solution: Given $\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sqrt{3} \sin x-\cos x}{x-\frac{\pi}{6}}$ Consider $\sqrt{3} \sin x-\cos x=2\left(\frac{\sqrt{3} \sin x}{2}-\frac{\cos x}{2}\right)=2\left(\sin x \cos \left(\frac{\pi}{6}\right)-\cos x \sin \left(\frac{\pi}{6}\right)\right)$ On simplification the above equation can be written as $\Rightarrow$$2\left(\sin x \cos...
$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}}$
[question] Question. $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}}$ [/question] [solution] solution: Given $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}}$ We have $\sin x-\cos x=\sqrt{2}\left(\frac{\sin x}{\sqrt{2}}-\frac{\cos x}{\sqrt{2}}\right)=\sqrt{2}\left(\sin x \cos \left(\frac{\pi}{4}\right)-\cos x \sin \left(\frac{\pi}{4}\right)\right)$ By using this formula in given equation we get $\Rightarrow \sqrt{2}\left(\sin x \cos \left(\fra...
$\lim _{x \rightarrow \frac{\pi}{3}} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2} \frac{\pi}{3}-x}$
[question] Question. $\lim _{x \rightarrow \frac{\pi}{3}} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2} \frac{\pi}{3}-x}$ [/question] [solution] solution: Given $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}$ Now by using the formula $\cos 6 x=1-2 \sin ^{2} 3 x$ Then the above equation becomes, $\Rightarrow$$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2}\left(\frac{\pi}{3}-x\right)}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sqrt{1-...
$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}$
[question] Question. $\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}$ [/question] [solution] solution: Given $\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}$ Here cos mx can be written as $\Rightarrow \cos m x=1-2 \sin ^{2} \frac{m x}{2}$ And similarly $\Rightarrow \operatorname{cosn} x=1-2 \sin ^{2} \frac{n x}{2}$ Using these two identities in given equation we get $\Rightarrow$$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0} \frac{\left[1-\left(1-2 \...
$\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{3}}$
[question] Question. $\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{3}}$ [/question] [solution] solution: Given $\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^{2}}$ We know that $\sin 2 x=2 \sin x \cos x$, using this formula we get Again by taking $2 \sin x$ common we get $\lim _{x \rightarrow 0} \frac{2 \sin x-2 \sin x \cos x}{x^{2}}=\lim _{x \rightarrow 0} \frac{2 \sin x(1-\cos x)}{x^{3}}$ Now we have $\cos x=1-2 \sin ^{2}(x / 2)$ Using this identity above equation can be written as...