MATHS 2 Marks Questions

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

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

$=\frac{2\text{apx}^3+\text{2aqx}^2+\text{2axr}+\text{bpx}^2+\text{bqx}+\text{br}-(\text{2apx}^3+\text{2pbx}^2+\text{2pcx}+\text{qax}^2+\text{bqx}+\text{cq})}{(\text{px}^2+\text{qx}+\text{r})^2}$

$=\frac{\text{2apx}^3-\text{2apx}^3+\text{2aqx}^2+\text{bpx}^2-\text{2pbx}^2-\text{qax}^2+\text{2arx}+\text{bqx}-\text{2pcx}-\text{bqx}+\text{br}-\text{cq}}{(\text{px}^2+\text{qx}+\text{r})^2}$

$=\frac{\text{aqx}^2-\text{bpx}^2+\text{2axr}-\text{2cpx}+\text{br}-\text{cq}}{(\text{px}^2+\text{qx}+\text{r})^2}$

$\frac{\text{x}^2(\text{aq}-\text{bp})+2(\text{ar}-\text{cp})\text{x}+\text{br}-\text{cq}}{(\text{px}^2+\text{qx}+\text{r})^2}$

$\frac{(\text{aq}-\text{bp})\text{x}^2+2(\text{ar}-\text{cp})\text{x}+\text{br}-\text{cq}}{(\text{px}^2+\text{qx}+\text{r})^2}$

Therefore,

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

$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}\Big(\frac{\pi}{2}+\text{h}\Big)-\text{f}\Big(\frac{\pi}{2}\Big)}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\Big(\frac{\pi}{2}+\text{h}\Big)-\sin2\Big(\frac{\pi}{2}\Big)}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Big(\frac{\pi}{2}\times2+2\text{h}\Big)-\sin(\pi)}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{-\cos2\text{h}-0}{\text{h}}$

$=-2$

Therefore, $\text{f}'\Big(\frac{\pi}{2}\Big)=-2$

$\frac{\text{d}}{\text{dx}}(\log_3\text{x}+3\log_\text{e}\text{x}+2\tan\text{x})$

$=\frac{1}{\log3}\frac{\text{d}}{\text{dx}}(\log\text{x})+3\frac{\text{d}}{\text{dx}}(\log_\text{e}\text{x})+2\frac{\text{d}}{\text{dx}}(\tan\text{x})\ \Bigg[\because\log_3\text{x}=\frac{\log\text{x}}{\log3}\Bigg]$

$=\frac{1}{\log3}\times\frac{1}{\text{x}}+\frac{3}{\text{x}}+2\sec^2\text{x}$

$=\frac{1}{\text{x}\log3}+\frac{3}{\text{x}}+2\sec^2\text{x}$

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

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

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

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

$\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\sin\text{x}}+2^{\text{x}+3}+\frac{4}{\log\text{x}^3}\Big)$

$=\frac{\text{d}}{\text{dx}}\text{cosec}\text{x}+2^2\frac{\text{d}}{\text{dx}}(2^{\text{x}})+\frac{4}{\log3}\times\frac{\text{d}}{\text{dx}}(\log\text{x})\ \Big[\because\log_\text{b}\text{a}=\frac{\log\text{a}}{\log\text{b}}\Big]$

$=-\text{coesc}\text{x}.\cot\text{x}+8.2^\text{x}\log2+\frac{4}{\log3}\times\frac{1}{\text{x}}\ \Big[\because\frac{\text{d}}{\text{dx}}(\text{a}^\text{x})=\text{a}^\text{x}\log\text{a}\Big]$

$=-\text{coesc}\text{x}.\cot\text{x}+2^{\text{x}+3}\log2+\frac{4}{\text{x}\log3}$

Then, $\text{u}'=\text{nx}^{\text{n}-1};\text{v}'=\frac{1}{\text{x}\log\text{a}}$

Using the product rule:

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

$\frac{\text{d}}{\text{dx}}(\text{x}^\text{n}\log_\text{a}\text{x})=\text{x}^\text{n}.\frac{1}{\text{x}\log\text{a}}+\log_\text{a}\text{x}(\text{nx}^{\text{n}-1})$

$=\text{x}^{\text{n}-1}\frac{1}{\log\text{a}}+\log_\text{a}\text{x}(\text{nx}^{\text{n}-1})$

$=\text{x}^{\text{n}-1}\Big(\frac{1}{\log\text{a}}+\text{n}\log_{\text{a}}\text{x}\Big)$

$=\frac{(\cot\text{x}-\text{x}^\text{n})\frac{\text{d}}{\text{dx}}(\text{e}^\text{x}-\tan\text{x})-(\text{e}^\text{x}-\tan\text{x})\frac{\text{d}}{\text{dx}}(\cot\text{x}-\text{x}^\text{n})}{(\cot\text{x}-\text{x}^\text{n})^2}$

$=\frac{(\cot\text{x}-\text{x}^\text{n})(\text{e}^\text{x}-\sec^2\text{x})-(\text{e}^\text{x}-\tan\text{x})(-\text{cosec}^2\text{x}-\text{nx}^{\text{n}-1})}{(\cot\text{x}-\text{x}^\text{n})^2}$

$=\frac{(\cot\text{x}-\text{x}^\text{n})(\text{e}^\text{x}-\sec^2\text{x})+(\text{e}^\text{x}-\tan\text{x})(\text{cosec}^2\text{x}-\text{nx}^{\text{n}-1})}{(\cot\text{x}-\text{x}^\text{n})^2}$

$=2\sin\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{x})$ (Using the chain rule)

$=2\sin\text{x}\cos\text{x}$

$=\sin\text{2x}$

$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{e}^\text{x}+\sin\text{x}}{1+\log\text{x}}\Big)$

Using Quotient rule, we get 

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

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

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

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

Using the product rule:

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

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

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

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

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

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

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

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

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

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

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

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

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

Then, $\text{u}'=3\text{x}^2;\text{v}'=\cos\text{x}$

Using the product rule:

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

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

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

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

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

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

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

$\Rightarrow\text{2x}\frac{\text{dy}}{\text{dx}}=\sqrt{\frac{\text{x}}{\text{a}}}-\sqrt{\frac{\text{a}}{\text{x}}}$

Mulitiplying both the side by $\text{y}=\sqrt{\frac{\text{x}}{\text{a}}}+\sqrt{\frac{\text{a}}{\text{x}}}$

$\Rightarrow\text{2xy}\frac{\text{dy}}{\text{dx}}=\Big(\sqrt{\frac{\text{x}}{\text{a}}}-\sqrt{\frac{\text{a}}{\text{x}}}\Big)\Big(\sqrt{\frac{\text{x}}{\text{a}}}+\sqrt{\frac{\text{a}}{\text{x}}}\Big)$

$=\Big(\frac{\text{x}}{\text{a}}-\frac{\text{a}}{\text{x}}\Big)$

Hence, proved.

$=\frac{(\text{x}+1)\times\text{2x}-(\text{x}^2+1)\times1}{(\text{x}+1)^2}$

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

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

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

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

$=4\text{x}^3-2\cos\text{x}-3\sin\text{x}$

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

Using the product rule:

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

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

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

$=\cos2\text{x}$

$\frac{\text{d}}{\text{dx}}(2\sec\text{x}+3\cot\text{x}-4\tan\text{x})$

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

$=2\sec\text{x}\tan\text{x}-3\text{cosec}^2\text{x}-4\sec^2\text{x}$

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

Using Quotient rule, we get 

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

$=\frac{(\sec\text{x}+\tan\text{x})(\text{x}\sec^2\text{x}+\tan\text{x})-(\text{x}\tan\text{x})(\sec\text{x}\tan\text{x}+\sec^2\text{x})}{(\sec\text{x}+\tan\text{x})^2}$ [Used product rule]

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

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

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

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

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

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

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

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

$=\frac{1}{3}.3\text{x}^2-2\frac{1.1}{2\sqrt{\text{x}}}+5.(-2)\text{x}^{-3}$

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

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

Then, $\text{u}'=2\text{x};\text{v}'=\cos\text{x};\text{w}'=\frac{1}{\text{x}}$

Using the product rule:

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

$\frac{\text{d}}{\text{dx}}(\text{x}^2\sin\text{x}\log\text{x})=2\text{x}\sin\text{x}\log\text{x}+\text{x}^2\cos\text{x}\log\text{x}+\text{x}^2\sin\text{x}.\frac{1}{\text{x}}$

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

Then, $\text{u}'=\text{nx}^{\text{n}-1};\text{v}'=\sec^2\text{x}$

Using the product rule:

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

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

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

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

$=\frac{-(\text{2ax}+\text{b})}{(\text{ax}^2+\text{bx}+\text{c})^2}$

$\therefore\frac{\text{d}}{\text{dx}}\frac{1}{\text{ax}^2+\text{bx}+\text{c}}=\frac{-(\text{2ax}+\text{b})}{(\text{ax}^2+\text{bx}+\text{c})^2}$

$=\text{x}^5\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})+\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^5)+(\text{x})^6\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^6)$

$=\text{x}^5\text{e}^\text{x}+\text{e}^\text{x}(5\text{x}^4)+\text{x}^6.\frac{1}{\text{x}}+\log\text{x}(6\text{x}^5)$

$=\text{x}^5\text{e}^\text{x}+\text{e}^\text{x}(5\text{x}^4)+\text{x}^5+\log\text{x}(6\text{x}^5)$

$=\text{x}^4(\text{x}\text{e}^\text{x}+5\text{e}^\text{x}+\text{x}+\text{6x}+\log\text{x})$

Then, $\text{u}'=\text{3x}^2+\text{2x};\text{v}'=\cos\text{x}$

Using the product rule:

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

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

$=(\text{x}^3+\text{x}^2+1)\cos\text{x}+(\text{3x}^2+\text{2x})\sin\text{x}$

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

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

$=3^\text{x}\log3+3\text{x}^2+0\ \Bigg[\because\frac{\text{d}}{\text{dx}}(\text{a}^\text{x})=\text{a}^\text{x}\log\text{a}\Bigg]$

$=3^\text{x}\log3+3\text{x}^2$

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

$=(3\text{x}+2)\frac{\text{d}}{\text{dx}}(2\text{x}^2+1)+(\text{2x}^2+1)\frac{\text{d}}{\text{dx}}(3\text{x}+2)$  [Using product rule]

$=(3\text{x}+2)(4\text{x}+0)+(\text{2x}^2+1)(3+0)$

$=(12\text{x}^2+\text{8x}+\text{6x}^2+3)$

$=18\text{x}^2+\text{8x}+3$

$\frac{\text{d}}{\text{dx}}(2\sec\text{x}+3\cot\text{x}-4\tan\text{x})$

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

$=2\sec\text{x}\tan\text{x}-3\text{cosec}^2\text{x}-4\sec^2\text{x}$

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

$=\text{a}\frac{\text{d}}{\text{dx}}\Big(\frac{\cos\text{x}}{\sin\text{x}}\Big)+\text{b}\frac{\text{d}}{\text{dx}}(1)+\text{c}\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\sin\text{x}}\Big)$

$=\text{a}(-\text{cosec}^2\text{x})+0+\text{c}(-\text{cosec}\text{x}.\cot\text{x})$

$=-\text{a}\text{cosesc}^2\text{x}-\text{c}\text{cosec}\text{x}.\cot\text{x}$

$\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}\log\text{a}}+\text{e}^{\text{a}\log\text{x}}+{\text{e}^{\text{a}\log\text{a}}})$

$=\frac{\text{d}}{\text{dx}}(\text{e}^{\text{x}\log\text{a}})+\frac{\text{d}}{\text{dx}}(\text{e}^{\text{a}\log\text{x}})+\frac{\text{d}}{\text{dx}}(\text{e}^{\text{a}\log\text{a}})$

$=\text{e}^{\text{x}\log\text{a}}.\log\text{a}+\text{e}^{\text{a}\log\text{x}}.\frac{\text{a}}{\text{x}}+0\ [\because\text{e}^{\text{a}\log\text{a}}$is constant$]$

$=\log\text{a}.\text{e}^{\text{x}\log\text{a}}+\frac{\text{a}}{\text{x}}\text{e}^{\text{a}\log\text{x}}$

$=\log\text{a}.\text{a}^\text{x}+\frac{\text{a}}{\text{x}}\text{x}^\text{a}\ [\text{a}^\text{x}$can be written as $\text{e}^{\text{x}\log\text{a}}]$

$=\text{a}^\text{x}\log\text{a}+\text{ax}^{\text{a}-1}$

$\text{f}'(0)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\cos(0+\text{h})-\cos0}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\cos\text{h}-\cos0}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\frac{\Big(1-\frac{\text{h}^2}{2!}+\frac{\text{h}^4}{4!}\dots\Big)-1}{\text{h}}\ \Big[\because\cos\text{x}=1-\frac{\text{x}^2}{2}+\frac{\text{x}^4}{4!}-\dots\Big]$

$\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}\Big(-\frac{\text{h}}{2!}+\frac{\text{h}^3}{4!}-\frac{\text{h}^5}{6!}-\dots\Big)}{\text{h}}$

$=\lim_\limits{\text{h}\rightarrow0}\text{h}\Big(-\frac{\text{h}}{2!}+\frac{\text{h}^3}{4!}-\frac{\text{h}^5}{6!}-\dots\Big)$

= 0

$\therefore\text{f}'(0)=0$

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

Using the product rule:

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

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

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

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

$\therefore\text{f}'\Big(\frac{\pi}{2}\Big)=0$

$\therefore\text{f}'(0)=1$