MATHS 3 Marks Question

$\text{f(x)}=\text{x}^3-5\text{x}^2+\text{x}-5$

$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\big(\text{x+h}\big)-\text{f}\big(\text{x}\big)}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\bigg\{(\text{x+h})^3+(\text{x+h})-5(\text{x+h})^2-5\bigg\}-(\text{x}^3-5\text{x}^2+\text{x}-5)}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{(\text{x}^3+\text{h}^3+3\text{x}^2\text{h}+3\text{h}^2\text{x}+\text{x}+\text{h}-5\text{x}^2-5\text{h}-10\text{xh}-5)-(\text{x}^3-5\text{x}^2+\text{x}-5)}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{(3\text{x}^2\text{h}+3\text{h}^2\text{x}+\text{h}^3+\text{h}-5\text{h}^2-10\text{xh})}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}3\text{x}^2+3\text{xh}+\text{h}^2+1-5\text{h}-10\text{x}$

$=3\text{x}^2-10\text{x}+1$

$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{\sqrt{\text{x+h}}}-\frac{1}{\sqrt{\text{x}}}}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{\text{x}}-\sqrt{\text{x+h}}}{\sqrt{\text{x}}\sqrt{\text{x+h}}.\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{\text{x}}-\sqrt{\text{x+h}}}{\sqrt{\text{x}}\sqrt{\text{x+h}}.\text{h}}\times\frac{\sqrt{\text{x}}+\sqrt{\text{x+h}}}{\sqrt{\text{x}}+\sqrt{\text{x+h}}.\text{h}}$

$$$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}-\text{(x+h)}}{\sqrt{\text{x}}\sqrt{\text{x+h}}.\text{h}\big(\sqrt{\text{x}}+\sqrt{\text{x+h}}\big)}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\sqrt{\text{x}}\sqrt{\text{x+h}}.\text{h}\big(\sqrt{\text{x}}+\sqrt{\text{x+h}}\big)}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\text{x}\sqrt{\text{x+h}}+\sqrt{\text{x}}\big(\sqrt{\text{x+h}}\big)}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{2\text{x}\sqrt{\text{x}}}$

$=\frac{-1}{2}\text{x}^{-\frac{3}{2}}$$$

$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2-\text{x}+1}{\text{x}^2+\text{x}+1}\Big)$

Using quotient rule, we get

$\frac{(\text{x}^2+\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}+1)-(\text{x}^2-\text{x}+1)\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{x}+1)}{(\text{x}^2+\text{x}+1)^2}$

$=\frac{(\text{x}^2+\text{x}+1)(\text{2x}-1)-(\text{x}^2-\text{x}+1)(\text{2x}+1)}{(\text{x}^2+\text{x}+1)^2}$

$=\frac{(\text{x}^2+1-\text{x})(\text{2x}-1)-(\text{x}^2-\text{x}+1)(\text{2x}+1)}{(\text{x}^2+\text{x}+1)^2}$

$=\frac{\text{2x}^3+\text{2x}+\text{2x}^2-\text{x}^2-1-\text{x}-\text{2x}^3+\text{2x}^2-\text{2x}-\text{x}^2+\text{x}-1}{(\text{x}^2+\text{x}+1)^2}$

$=\frac{\text{2x}^2-2}{(\text{x}^2+\text{x}+1)^2}$

$=\frac{2(\text{x}^2-1)}{(\text{x}^2+\text{x}+1)^2}$

Multiplying Numerator and Denominator by 3

$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{1+\tan\text{x}}\Big)$

Using quotient rule, we get

$\frac{(1+\tan\text{x})\frac{\text{d}}{\text{dx}}(\text{x})-(\text{x})\frac{\text{d}}{\text{dx}}(1+\tan\text{x})}{(1+\tan\text{x})^2}$

$=\frac{(1+\tan\text{x})-\text{x}\sec^2\text{x}}{(1+\tan\text{x})^2}$

$=\frac{1+\tan\text{x}-\text{x}\sec^2\text{x}}{(1+\tan\text{x})^2}$

Then, $\text{u}'=\text{e}^\text{x};\text{v}'=\frac{1}{\sqrt{\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}=\frac{1}{\text{2x}};\text{w}'=\sec^2\text{x}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uvw})=\text{u}'\text{vw}+\text{uv}'\text{w}+\text{uv}\text{w}'$

$=\text{e}^{\text{x}}\log\sqrt{\text{x}}\tan\text{x}+\text{e}^\text{x}\times\frac{1}{\text{2x}}\tan\text{x}+\text{e}^\text{x}\log\sqrt{\text{x}}\sec^2\text{x}$

$=\text{e}^\text{x}(\log\text{x}^{\frac{1}{2}}.\tan\text{x}+\frac{\tan\text{x}}{\text{2x}}+\log\text{x}^{\frac{1}{2}}.\sec^2\text{x})$

$=\text{e}^\text{x}\Big(\frac{1}{2}\log\text{x}.\tan\text{x}+\frac{\tan\text{x}}{\text{2x}}+\frac{1}{2}\log\text{x}.\sec^2\text{x}\Big)$

$=\frac{\text{e}^\text{x}}{2}\Big(\log\text{x}.\tan\text{x}+\frac{\tan\text{x}}{\text{2x}}+\log\text{x}.\sec^2\text{x}\Big)$

$\frac{\text{d}}{\text{dx}}\Big(\frac{\sec\text{x}-1}{\sec\text{x}+1}\Big)$

Using quotient rule, we get

$\frac{(\sec\text{x}+1)\frac{\text{d}}{\text{dx}}(\sec\text{x}-1)-(\sec\text{x}-1)\frac{\text{d}}{\text{dx}}(\sec\text{x}+1)}{(\sec\text{x}+1)^2}$

$=\frac{(\sec\text{x}+1)(\sec\text{x}\tan\text{x})-(\sec\text{x}-1)(\sec\text{x}\tan\text{x})}{(\sec\text{x}+1)^2}$

$=\frac{\sec\text{x}\tan\text{x}(\sec\text{x}+1-\sec\text{x}+1)}{(\sec\text{x}+1)^2}$

$=\frac{2\sec\text{x}\tan\text{x}}{(\sec\text{x}+1)^2}$

Let $\text{u}=\text{x}^2;\text{v}=\cos\frac{\pi}{4};\text{w}=\text{cosec}\text{x}$

Then, $\text{u}'=\text{2x};\text{v}'=0;\text{w}'=-\text{cosecx}\cot\text{x}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uvw})=\text{u}'\text{vw}+\text{uv}'\text{w}+\text{uv}\text{w}'$

$\frac{\text{d}}{\text{dx}}(\text{x}^2\cos\frac{\pi}{4}\text{cosec}\text{x})=\text{2x}\cos\frac{\pi}{4}\text{cosec}\text{x}+\text{x}^2.0.\text{\cosecx}+\text{x}^2\cos\frac{\pi}{4}(-\text{cosecx}\cot\text{x})$

$=\cos\frac{\pi}{4}(\text{2x}\text{cosecx}-\text{x}^2\text{cosecx}\cot\text{x})$

$=\cos\frac{\pi}{4}\Big(\frac{\text{2x}}{\sin\text{x}}-\text{x}^2\frac{\cot\text{x}}{\sin\text{x}}\Big)$

Then, $\text{u}'=-2\sec^2\text{x};\text{v}'=4\cos\text{x}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$

$\frac{\text{d}}{\text{dx}}=[(1-2\tan\text{x})(5+4\sin\text{x})​​]$

$=(1-2\tan\text{x})(4\cos\text{x})+(5+4\sin\text{x})(-2\sec^2\text{x})$

$=4\cos\text{x}-8\times\frac{\sin\text{x}}{\cos\text{x}}\cos\text{x}-10\sec^2\text{x}-8\frac{\sin\text{x}}{\cos^2\text{x}}$

$=4\cos\text{x}-8\sin\text{x}-10\sec^2\text{x}-8\sec\text{x}\tan\text{x}$

$=4\Big(\cos\text{x}-2\sin\text{x}-\frac{5}{2}\sec^2\text{x}-2\sec\text{x}\tan\text{x}\Big)$

$\text{f}(\text{x})=\text{x}^4-\text{2x}^3+\text{3x}^2+\text{x}+5$

Differentiate with respect to $\text{x}$

$\frac{\text{d}}{\text{dx}}\text{f}(\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}^4-\text{2x}^3+\text{3x}^2+\text{x}+5)$

$=\text{4x}^3-\text{6x}^2+\text{6x}+1$

Then, $\text{u}'=\text{6x};\text{v}'=\text{6x}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$

$\frac{\text{d}}{\text{dx}}[(\text{3x}^2+2)(\text{3x}^2+2)]$

$=(\text{3x}^2+2)(\text{6x})+(\text{3x}^2+2)(\text{6x})$

$=18\text{x}^3+\text{12x}+\text{18x}^3+\text{12x}$

$=\text{36x}^3+\text{24x}$

Alternate method

$\frac{\text{d}}{\text{dx}}\Big[(\text{3x}^2+2)^2\Big]=\frac{\text{d}}{\text{dx}}(\text{9x}^4+\text{12x}^2+4)$

$=\text{36x}^3+\text{24x}$

$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{2-3\cos\text{x}}{\sin\text{x}}\Big)$

$=\frac{\text{d}}{\text{dx}}(2\text{cosec}\text{x}-3\cot\text{x})$

$=2\frac{\text{d}}{\text{dx}}(\text{cosec}\text{x})-3\frac{\text{d}}{\text{dx}}(\cot\text{x})$

$=-2\text{cosec}\text{x}.\cot\text{x}+3\text{cosec}^2\text{x}$

$\frac{\text{dy}}{\text{dx}}$ at $\text{x}=\frac{\pi}{4}$

$=-2\text{cosec}\frac{\pi}{4}.\cot\frac{\pi}{4}+3\text{cosec}^2\frac{\pi}{4}$

$=-2\sqrt2-1+3.2$

$=-2\sqrt2+6$

$=6-2\sqrt2$

Let $\text{u}=(\text{ax}+\text{b})^\text{n},\text{v}=(\text{cx}+\text{d})^\text{m}$

Then, $\text{u}'=\text{na}(\text{ax}+\text{b})^{\text{n}-1},\text{v}'=\text{nc}(\text{cx}+\text{d})^{\text{m}-1}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$

$\frac{\text{d}}{\text{dx}}[(\text{ax}+\text{b})^\text{n}(\text{cx}+\text{d})^\text{m}]=(\text{ax}+\text{b})^\text{n}\times\text{mc}(\text{cx}+\text{d})^{\text{m}-1}+(\text{cx}+\text{d})^\text{m}\times\text{na}(\text{ax}+\text{b})^{\text{n}-1}$

$\text{mc}(\text{ax}+\text{b})^\text{n}(\text{cx}+\text{d})^{\text{m}-1}+\text{na}(\text{cx}+\text{d})^\text{m}(\text{ax}+\text{b})^{\text{n}-1}$

$\frac{\text{d}}{\text{dx}}\Big\{\cos(\text{x}+\text{h})\Big\}$

$=\frac{\text{d}}{\text{dx}}(\cos\text{x}.\cos\text{a}-\sin\text{x}.\sin\text{a})\ [\because\cos(\text{x}+\text{a})=\cos\text{x}\cos\text{a}-\sin\text{x}\sin\text{a}]$

$=\cos\text{a}\frac{\text{d}}{\text{dx}}(\cos\text{x})-\sin\text{a}\frac{\text{d}}{\text{dx}}(\sin\text{x})$

$=\cos\text{a}(-\sin\text{x})-\sin\text{a}(\cos\text{x})$

$=\cos\text{x}(\sin\text{a})+\sin\text{x}(\cos\text{a})$

$=-(\sin\text{x}\cos\text{a}+\cos\text{x}\sin\text{a})$

$=-\sin(\text{x}+\text{a})$

$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{px}^2+\text{qx}+\text{r}}{\text{ax}+\text{b}}\Big)$

Using quotient rule, we get

$\frac{(\text{ax}+\text{b})\frac{\text{d}}{\text{dx}}(\text{px}^2+\text{qx}+\text{r})-(\text{px}^2+\text{qx}+\text{r})\frac{\text{d}}{\text{dx}}(\text{ax}+\text{b})}{(\text{ax}+\text{b})^2}$

$=\frac{(\text{ax}+\text{b})(\text{2px}+\text{q})-(\text{px}^2\text{qx}+\text{r})\text{a}}{(\text{ax}+\text{b})^2}$

$=\frac{\text{2apx}^2+\text{2pbx}+\text{aqx}+\text{bq}-\text{apx}^2-\text{aqx}-\text{ar}}{(\text{ax}+\text{b})^2}$

$=\frac{\text{apx}^2+\text{2pbx}+\text{bq}-\text{ar}}{(\text{ax}+\text{b})^2}$

$\frac{\text{d}}{\text{dx}}(\text{a}_0\text{x}^\text{n}+\text{a}_1\text{x}^{\text{n}-1}+\text{a}_2\text{x}^{\text{n}-2}+\dots+\text{a}_{\text{n}-1}+\text{a}_\text{n})$

$=\text{a}_0\frac{\text{d}}{\text{dx}}(\text{x})^\text{n}+\text{a}_1\frac{\text{d}}{\text{dx}}(\text{x})^{\text{n}-1}+\text{a}_2\frac{\text{d}}{\text{dx}}(\text{x})^{\text{n}-2}+\dots+\text{a}_{\text{n}-1}\frac{\text{d}}{\text{dx}}+\text{a}_\text{n}\frac{\text{d}}{\text{dx}}(1)$

$=\text{na}_0\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_1\text{x}^{\text{n}-2}+\dots+\text{a}_{\text{n}-1}+0$

$=\text{na}_0\text{x}^{\text{n}-1}+(\text{n}-1)\text{a}_1\text{x}^{\text{n}-2}+\dots+\text{a}_{\text{n}-1}$

$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^\text{n}}{\sin\text{x}}\Big)$

$=\text{x}^\text{n}\frac{\text{d}}{\text{dx}}(\sin\text{x})^{-1}+\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}^\text{n})$

$=\text{x}^\text{n}\frac{-1}{\sin^2\text{x}}+\frac{1}{\sin\text{x}}\text{n}\text{x}^{\text{n}-1}$

$=\frac{\sin\text{x}(\text{nx}^{\text{n}-1})-\text{x}^\text{n}(\cos\text{x})}{\sin^2\text{x}}$

Then, $\text{u}'=-\text{3x}^{-4};\text{v}'=3$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$

$\frac{\text{d}}{\text{dx}}[\text{x}^{-3}(5+\text{3x})]=\text{x}^{-3}.3+(5+\text{3x})(-3\text{x}^{-4})$

$=\text{3x}^{-3}-\text{15x}^{-4}-\text{9x}^{-3}$

$=-\text{15x}^{-4}-\text{6x}^{-3}$

$\frac{\text{d}}{\text{dx}}\frac{(\text{x}^3+1)(\text{x}-2)}{\text{x}^2}$

$=\frac{\text{d}}{\text{dx}}\frac{(\text{x}^4-\text{2x}^3+\text{x}-2)}{\text{x}^2}$

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

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

$=\text{2x}-2-\frac{1}{\text{x}^2}+2.\frac{2}{\text{x}^3}$

$=\text{2x}-2-\frac{1}{\text{x}^2}+\frac{4}{\text{x}^3}$

Then, $\text{u}'=\text{2x};\text{v}'=-\sin\text{x}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$

$\frac{\text{d}}{\text{dx}}=[(1+\text{x}^2)\cos\text{x}]=(1+\text{x}^2)(-\sin\text{x})+(\cos\text{x})(\text{2x})$

$=-\sin\text{x}-\text{x}^2\sin\text{x}+\text{2x}\cos\text{x}$

$\frac{\text{d}}{\text{dx}}\Big(\frac{2\text{x}^2+\text{3x}+4}{\text{x}}\Big)$

$=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{2x}^2}{\text{x}}+\frac{\text{3x}}{\text{x}}+\frac{4}{\text{x}}\Big)$

$=\frac{\text{d}}{\text{dx}}(\text{2x}+3+4\text{x}^{-1})$

$=2-\frac{4}{\text{x}^2}$

$\frac{\text{d}}{\text{dx}}\Big(\frac{\sqrt{\text{a}}+\sqrt{\text{x}}}{\sqrt{\text{a}}-\sqrt{\text{x}}}\Big)$

Using quotient rule, we get

$\frac{(\sqrt{\text{a}}-\sqrt{\text{x}})\frac{\text{d}}{\text{dx}}(\sqrt{\text{a}}+\sqrt{\text{x}})-(\sqrt{\text{a}}+\sqrt{\text{x}})\frac{\text{d}}{\text{dx}}(\sqrt{\text{a}}-\sqrt{\text{x}})}{(\sqrt{\text{a}}-\sqrt{\text{x}})^2}$

$=\frac{(\sqrt{\text{a}}-\sqrt{\text{x}})\times\frac{1}{2\sqrt{\text{x}}}-(\sqrt{\text{a}}+\sqrt{\text{x}})\times\frac{-\text{1}}{\text{2}\sqrt{\text{x}}}}{(\sqrt{\text{a}}-\sqrt{\text{x}})^2}$

$=\frac{\sqrt{\text{a}}-\sqrt{\text{x}}+\sqrt{\text{a}}+\sqrt{\text{x}}}{2\sqrt{\text{x}}(\sqrt{\text{a}}-\sqrt{\text{x}})^2}$

$=\frac{\sqrt{\text{a}}}{\sqrt{\text{x}}(\sqrt{\text{a}}-\sqrt{\text{x}})^2}$

$\frac{\text{d}}{\text{dx}}\Big(\frac{1+\log\text{x}}{1-\log\text{x}}\Big)$

Using quotient rule, we get

$\frac{(1-\log\text{x})\frac{\text{d}}{\text{dx}}(1+\log\text{x})-(1+\log\text{x})\frac{\text{d}}{\text{dx}}(1-\log\text{x})}{(1-\log\text{x})^2}$

$\frac{(1-\log\text{x})\times\frac{1}{\text{x}}-(1+\log\text{x})\big(\frac{-1}{\text{x}}\big)}{(1-\log\text{x})^2}$

$=\frac{1-\log\text{x}+1+\log\text{x}}{\text{x}(1-\log\text{x})^2}$

$=\frac{2}{\text{x}(1-\log\text{x})^2}$

Then, $\text{u}'=\text{3x}^2;\text{v}'=\text{e}^\text{x};\text{w}'=-\sin\text{x}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uvw})=\text{u}'\text{vw}+\text{uv}'\text{w}+\text{uv}\text{w}'$

$\frac{\text{d}}{\text{dx}}(\text{x}^3\text{e}^\text{x}\cos\text{x})=\text{3x}^2\text{e}^\text{x}\cos\text{x}+\text{x}^3\text{e}^\text{x}\cos\text{x}+\text{x}^3\text{e}^\text{x}(-\sin\text{x})$

$=\text{x}^2\text{e}^\text{x}(\text{3}\cos\text{x}+\text{x}\cos\text{x}-\text{x}\sin\text{x})$

$\frac{\text{d}}{\text{dx}}\frac{\cos(\text{x}-2)}{\sin\text{x}}$

$=\frac{\text{d}}{\text{dx}}\frac{(\cos\text{x}.\cos2+\sin\text{x}.\sin2)}{\sin\text{x}}$

$=\cos2\frac{\text{d}}{\text{dx}}(\cot\text{x})+\sin2\frac{\text{d}}{\text{dx}}(1)$

$-\cos2.\text{coesc}^2\text{x}+0$

$-\cos2.\text{coesc}^2\text{x}$

$\frac{\text{d}}{\text{dx}}\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$

$=\Big(\text{x}+\frac{1}{\text{x}}\Big)\frac{\text{d}}{\text{dx}}\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)+\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)\frac{\text{d}}{\text{dx}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$ [Using product rule]

$=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Bigg(\frac{1}{2\sqrt{\text{x}}}-\frac{1}{2\text{x}^{\frac{3}{2}}}\Bigg)+\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)\Big(1-\frac{1}{\text{x}^2}\Big)$

$=\Bigg(\frac{\text{x}}{2\sqrt{\text{x}}}-\frac{\text{x}}{\text{2x}^{\frac{3}{2}}}+\frac{1}{\text{2x}^{\frac{3}{2}}}-\frac{1}{\text{2x}^{\frac{5}{2}}}\Bigg)+\Bigg(\sqrt{\text{x}}-\frac{\sqrt{\text{x}}}{\text{x}^2}+\frac{1}{\sqrt{\text{x}}}-\frac{1}{\text{x}^{\frac{5}{2}}}\Bigg)$

$=\Bigg(\frac{1}{2}\sqrt{\text{x}}-\frac{1}{2\sqrt{\text{x}}}+\frac{1}{\text{2x}^{\frac{3}{2}}}-\frac{1}{\text{2x}^{\frac{5}{2}}}+\sqrt{\text{x}}-\frac{1}{\text{x}^{\frac{3}{2}}}+\frac{1}{\sqrt{\text{x}}}-\frac{1}{\text{x}^{\frac{5}{2}}}\Bigg)$

$=\Bigg(\frac{3}{2}\sqrt{\text{x}}+\frac{1}{2}\sqrt{\text{x}}-\frac{1}{\text{2x}^{\frac{3}{2}}}-\frac{3}{\text{2x}^{\frac{5}{2}}}\Bigg)$

$=\frac{3}{2}\text{x}^{\frac{1}{2}}+\frac{1}{2}\text{x}^{\frac{-1}{2}}-\frac{1}{2}\text{x}^{\frac{-3}{2}}-\frac{3}{2}\text{x}^{\frac{-5}{2}}$

$\frac{\text{d}}{\text{dx}}\Big(\frac{10^\text{x}}{\sin\text{x}}\Big)$

Using quotient rule, we get

$\frac{(\sin\text{x})\frac{\text{d}}{\text{dx}}(10^\text{x})-(10^\text{x})\frac{\text{d}}{\text{dx}}(\sin\text{x})}{(\sin\text{x})^2}$

$=\frac{\sin\text{x}\times10^\text{x}\log10-10^\text{x}\cos\text{x}}{(\sin\text{x})^2}$

$=10^\text{x}\text{cosecx}\log10-10^\text{x}\text{cosecx}\cot\text{x}$

$=10^\text{x}\text{cosecx}(\log10-\cot\text{x})$

Then, $\text{u}'=-\text{4x}^{-5};\text{v}'=\text{20x}^{-6}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$

$\frac{\text{d}}{\text{dx}}[\text{x}^{-4}(3-\text{4x}^{-5})]=\text{x}^{-4}(\text{20x}^{-6})+(3-\text{4x}^{-5})(-\text{4x}^{-5})$

$=\text{20x}^{-10}-\text{12x}^{-5}+\text{16x}^{-10}$

$=-\text{12x}^{-5}+\text{36x}^{-10}$

$\frac{\text{d}}{\text{dx}}\Big(\frac{3^\text{x}}{\text{x}+\tan\text{x}}\Big)$

Using quotient rule, we get

$\frac{(\text{x}+\tan\text{x})\frac{\text{d}}{\text{dx}}(3^\text{x})-(3^\text{x})\frac{\text{d}}{\text{dx}}(\text{x}+\tan\text{x})}{(\text{x}+\tan\text{x})^2}$

$=\frac{(\text{x}+\tan\text{x})\times3^\text{x}\log3-3^\text{x}(1+\sec^2\text{x})}{(\text{x}+\tan\text{x})^2}$

$=\frac{3^\text{x}\Big\{(\text{x}+\tan\text{x})\log3-(1+\sec^2\text{x})\Big\}}{(\text{x}+\tan\text{x})^2}$

$\frac{\text{d}}{\text{dx}}\Big\{\frac{(\text{x}+5)(\text{2x}^2-1)}{\text{x}}\Big\}$

$=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{2x}^3+\text{10x}^2-\text{x}-5}{\text{x}}\Big)$

$=\frac{\text{d}}{\text{dx}}(\text{2x}^2+\text{10x}-1-5\text{x}^{-1})$

$=2\frac{\text{d}}{\text{dx}}(\text{x}^2)+10\frac{\text{d}}{\text{dx}}(\text{x})-\frac{\text{d}}{\text{dx}}(1)-5\frac{\text{d}}{\text{dx}}(\text{x}^{-1})$

$=2\times2\text{x}+10-0+\frac{5}{\text{x}^2}$

$=4\text{x}+10+\frac{5}{\text{x}^2}$

Then, $\text{u}'=\text{4x};\text{v}'=\cos\text{x}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$

$\frac{\text{d}}{\text{dx}}[(\text{2x}^2-3)\sin\text{x}]=(\text{2x}^2-3)\cos\text{x}+\text{4x}\sin\text{x}$

$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{2x}^9}{3}-\frac{\text{5x}^7}{7}+\text{6x}^3-\text{x}\Big)$

$=\frac{2}{3}\frac{\text{d}}{\text{dx}}(\text{x})^9-\frac{5}{7}\frac{\text{d}}{\text{dx}}(\text{x})^7+6\frac{\text{d}}{\text{dx}}(\text{x})^3-\frac{\text{x}}{\text{dx}}(\text{x})$

$=\frac{2}{3}.9\text{x}^8-\frac{5}{7}.7\text{x}^6+\text{18x}^2-1$

$=\text{6x}^8-\text{5x}^6+\text{18x}^2-1$

$\therefore\frac{\text{dy}}{\text{dx}}$ at $\text{x}=1$

$=\text{6}(1)^8-\text{5}(1)^6+\text{18}(1)^2-1$

$=6-5+18-1$

$=18$

$\frac{\text{d}}{\text{dx}}\Big(\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)^3$

$=\frac{\text{d}}{\text{dx}}\Bigg[(\sqrt{\text{x}})^3+3(\sqrt{\text{x}})^2\Big(\frac{1}{\sqrt{\text{x}}}\Big)+3(\sqrt{\text{x}})\Big(\frac{1}{\sqrt{\text{x}}}\Big)^2+\Big(\frac{1}{\sqrt{\text{x}}}\Big)^3\Bigg]$

$=\frac{\text{d}}{\text{dx}}\big(\text{x}^{\frac{3}{2}}\big)+3\frac{\text{d}}{\text{dx}}\big(\text{x}^{\frac{1}{2}}\big)+3\frac{\text{d}}{\text{dx}}\big(\text{x}^{\frac{-1}{2}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{x}^{\frac{-3}{2}}\big)$

$=\frac{3}{2}\text{x}^{\frac{3}{2}-1}+3.\frac{1}{2}\text{x}^{\frac{1}{2}-1}+3\Big(\frac{-1}{2}\Big)\text{x}^{\frac{-1}{2}-1}+\Big(\frac{-3}{2}\Big)\text{x}^{\frac{-3}{2}-1}$

$=\frac{3}{2}\text{x}^\frac{1}{2}+\frac{3}{2}\text{x}^\frac{-1}{2}-\frac{3}{2}\text{x}^\frac{-3}{2}-\frac{3}{2}\text{x}^\frac{-5}{2}$

$\text{u}'=\text{x}\cos\text{x}+\sin\text{x}-\sin\text{x}=\text{x}\cos\text{x}$

$\text{v}'=\text{e}^\text{x}+\text{x}+\text{2x}\log\text{x}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$

$\frac{\text{d}}{\text{dx}}=[(\text{x}\sin\text{x}+\cos\text{x})(\text{e}^\text{x}+\text{x}^2\cos\text{x})]$

$=(\text{x}\sin\text{x}+\cos\text{x})(\text{e}^\text{x}+\text{x}+\text{2x}\log\text{x})+(\text{e}^\text{x}+\text{x}^2\log\text{x})(\text{x}\cos\text{x})$

$\frac{\text{d}}{\text{dx}}\Big(\frac{2^\text{x}\cot\text{x}}{\sqrt{\text{x}}}\Big)$

Using quotient rule, we get

$\frac{\sqrt{\text{x}}\frac{\text{d}}{\text{dx}}(2^\text{x}\cot\text{x})-(2^\text{x}\cot\text{x})\frac{\text{d}}{\text{dx}}(\sqrt{\text{x}})}{(\sqrt{\text{x}})^2}$

$=\frac{\sqrt{\text{x}}\Big(2^\text{x}\frac{\text{d}}{\text{dx}}\cot\text{x}+\cot\text{x}\frac{\text{d}}{\text{dx}}2^\text{x}\Big)-2^\text{x}\cot\text{x}\times\frac{1}{2}\text{x}^{-\frac{1}{2}}}{(\sqrt{\text{x}})^2}$

$=\frac{\sqrt{\text{x}}(2^\text{x}-\text{cosec}^2\text{x}+\cot\text{x}\times\log2\times2^\text{x})-2^\text{x}\cot\text{x}\times\frac{1}{2\sqrt{\text{x}}}}{(\sqrt{\text{x}})^2}$

$=\frac{2^\text{x}\Big\{-\text{x}\text{cosec}^2\text{x}+\text{x}\cot\text{x}\times\log2-\Big(\frac{1}{2}\Big)\cot\text{x}\Big\}}{(\sqrt{\text{x}})^2\times\sqrt{\text{x}}}$

$=\frac{2^\text{x}\Big(-\text{x}\text{cosec}^2\text{x}+\text{x}\cot\text{x}\times\log2-\Big(\frac{1}{2}\Big)\cot\text{x}\Big)}{\text{x}^{\frac{3}{2}}}$

Then, $\text{u}'=\text{5x}^4;\text{v}'=\text{54x}^{-10}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$

$\frac{\text{d}}{\text{dx}}[\text{x}^5(3-\text{6x}^9)]=\text{x}^5(\text{54x}^{-10})+\text{5x}^4(3-\text{6x}^9)$

$=\text{54x}^{-5}+\text{15x}^4-\text{30x}^{-5}$

$=\text{15x}^4+\text{24x}^{-5}$

$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{a}+\text{b}\sin\text{x}}{\text{c}+\text{d}\cos\text{x}}\Big)$

Using quotient rule, we get

$\frac{(\text{c}+\text{d}\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{a}+\text{b}\sin\text{x})-(\text{a}+\text{b}\sin\text{x})\frac{\text{d}}{\text{dx}}(\text{c}+\text{d}\cos\text{x})}{(\text{c}+\text{d}\cos\text{x})^2}$

$=\frac{(\text{c}+\text{d}\cos\text{x})(\text{b}\cos\text{x})-(\text{a}+\text{b}\sin\text{x})(-\text{d}\sin\text{x})}{(\text{c}+\text{d}\cos\text{x})^2}$

$=\frac{\text{bc}\cos\text{x}+\text{bd}\cos^2\text{x}+\text{ad}\sin\text{x}+\text{bd}\sin^2\text{x}}{(\text{c}+\text{d}\cos\text{x})^2}$

$=\frac{\text{bc}\cos\text{x}+\text{ad}\sin\text{x}+\text{bd}(\sin^2\text{x}+\cos^2\text{x})}{(\text{c}+\text{d}\cos\text{x})^2}$

$=\frac{\text{bc}\cos\text{x}+\text{ad}\sin\text{x}+\text{bd}}{(\text{c}+\text{d}\cos\text{x})^2}$

We have,

$\text{f}(\text{x})=\text{2x}^6+\text{x}^4-1$

$=\frac{\text{d}}{\text{dx}}(\text{2x}^6+\text{x}^4-1)$

$2\frac{\text{d}}{\text{dx}}(\text{x}^6)+\frac{\text{d}}{\text{dx}}(\text{x}^4)-\frac{\text{d}}{\text{dx}}(1)$

$=12\text{x}^5+4\text{x}^3-0$

$=12\text{x}^5+4\text{x}^3$

$\frac{\text{dy}}{\text{dx}}$ at $\text{x}=1$

$=12(1)^5+4(1)^3$

$=12+4$

$=16$

$\therefore$ The slope of the tangent to the curve $\text{f}(\text{x})=\text{2x}^6+\text{x}^4-1$ is 16.

$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\sin\frac{\text{x}}{2}+\cos\frac{\text{x}}{2}\Big)^2$

$=\frac{\text{d}}{\text{dx}}\Big(\sin^2\frac{\text{x}}{2}+\cos^2\frac{\text{x}}{2}+2\sin\frac{\text{x}}{2}.\cos\frac{\text{x}}2{}\Big)$

$=\frac{\text{d}}{\text{dx}}(1+\sin\text{x})\ \begin{bmatrix}\because\sin^2\theta+\cos^2\theta=1\\\sin^2\theta=2\sin\theta.\cos\theta\end{bmatrix}$

$=0+\cos\text{x}$

$=\cos\text{x}$

$\frac{\text{dy}}{\text{dx}}$ at $\text{x}=\frac{\pi}{6}$

$=\cos\frac{\pi}{6}$

$=\frac{\sqrt3}{2}$

$\text{f}(\text{x})=\text{e}^{\text{x}^2+1}$

$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{(\text{x}+\text{h})^2}+1-\text{e}^ {\text{x}^2+1}}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\text{x}^2+\text{h}^2+\text{2xh}+1}-\text{e}^ {\text{x}^2+1}}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\text{x}^2+1}(\text{e}^{\text{2xh}}.\text{e}^{\text{h}^2}-1)}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\text{x}^2+1}(\text{e}^{\text{2xh}+\text{h}^2}-1)}{\text{2xh}+\text{h}^2}\times\frac{\text{2xh}+\text{h}^2}{\text{h}}$

$\because\text{h}\rightarrow0$

$\Rightarrow\text{2xh}+\text{h}^2=0$

and $\lim_\limits{\theta\rightarrow0}\frac{\text{e}^\theta-1}{\theta}=1$

$=\lim_\limits{\text{h}\rightarrow0}\text{e}^{\text{x}^2+1}.1\times\text{2x}+\text{h}$

$=\text{2xe}^{\text{x}^2+1}$

$\text{f}(\text{x})=\text{e}^{\sqrt{2\text{x}}}$

$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{2(\text{x}+\text{h})}}-\text{e}^{\sqrt{2\text{x}}}}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{2\text{x}}}\big(\text{e}^{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}-1\big)}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{e}^{\sqrt{2\text{x}}}\big(\text{e}^{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}-1\big)}{\sqrt{2(\text{x}+\text{h})-\sqrt{2\text{x}}}}\times\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{\text{h}}$

Multiplying numerator and denominator by $\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}$

$\because\text{h}\rightarrow0\Rightarrow\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}\Rightarrow0$

and  $\lim_\limits{\theta\rightarrow0}\frac{\text{e}^\theta-1}{\theta}=1$

$\therefore$$\lim_\limits{\text{h}\rightarrow0}$$​​​​\text{e}^{\sqrt{2\text{x}}}$$\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{\text{h}}$

Again Multiplying numerator and denominator by $\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}$

$\therefore\lim_\limits{\text{h}\rightarrow0}\text{e}^{\sqrt{2\text{x}}}\times\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{\text{h}}\times\frac{\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}}{\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}}$

$=\text{e}^{\sqrt{2\text{x}}}\times\frac{1}{2\sqrt{2\text{x}}}$

$\text{f(x)}=\frac{\text{x+2}}{3\text{x}+5}$

$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{(x+h+2)}}{3\text{(x+h)+5}}-\frac{\text{x+2}}{3\text{x+5}}}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{3x+5}\big)\big(\text{x+h+2}\big)-\big(\text{x+2}\big)\big(\text{3x+3h+5}\big)}{\big(\text{3x+5}\big)\big(\text{3x+3h+5}\big)\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\big(\text{3x}^2+5\text{x}+3\text{xh}+5\text{h}+6\text{x}+10\big)-\big(\text{3x}^2+3\text{xh}+5\text{x}+6\text{x}+6\text{h}+10\big)}{\big(\text{3x+5}\big)\big(\text{3x+3h+5}\big)\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\big(\text{3x+5}\big)\big(\text{3x+3h+5}\big)\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\big(\text{3x+5}\big)\big(\text{3x+3h+5}\big)}$

$=\frac{-1}{\big(\text{3x+5}\big)^2}$

$\text{f(x)}=\text{x}^2+\text{x}+3$

$\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}\big(\text{x+h}\big)-\text{f}\big(\text{x}\big)}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\bigg\{{\big(\text{x+h}\big)^2+\big(\text{x+h}\big)+3}\bigg\}-\text{x}^2+\text{x}+3}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}^2+\text{h}^2+2\text{xh}+\text{x+h+3}-\text{x}^2-\text{x}-3}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{xh}+\text{h}^2+\text{h}}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}\big(2\text{x+h+1}\big)}{\text{h}}$

$=2\text{x}+0+1$

$=2\text{x}+1$