Differentiation EX 1.1 Q.2 Maharashtra state Board Mathematics and statistic for arts and science

Differentiation

Ex 1.1

Q.2

i) $$y=\cos \left(x^{2}+a^{2}\right)\\ Diff w.r.t.\quad x\\ \frac{d y}{d x}=\frac{d}{d x} \cos \left(x^{2}+a^{2}\right)\\ =-\sin \left(x^{2}+a^{2}\right) \cdot \frac{d}{d x}\left(x^{2}+a^{2}\right)\\ =-\sin \left(x^{2}+a^{2}\right)\left(\frac{d}{d x} x^{2}+\frac{d}{d x} a^{2}\right)\\ =-\sin \left(x^{2}+a^{2}\right)(2 x+0)\\ =-2 x \sin \left(x^{2}+a^{2}\right)\\$$

ii)  $$\quad y=\sqrt{e^{(3 x+2)}+5}\\ Diff. w.r.t \quad x\\ \frac{d y}{d x}=\frac{d}{d x} \sqrt{e^{(2 x+2)}+5}\\ =\frac{1}{2 \sqrt{e^{(3 x+2)}+5}} \cdot \frac{d}{d x}\left(e^{3 x+2)}+5\right)\\ =\frac{1}{2 \sqrt{e^{(3 x+2)}+5}} \cdot\left(\frac{d}{d x} e^{(3 x+2)}+\frac{d}{d x} 5\right)\\ =\frac{1}{2\sqrt{e^{3 x+2}+5}}\left(e^{(3 x+2)}\frac{d}{d x}(3 x+2)+0\right)\\ =\frac{1}{2\sqrt{e^{(3 x+2)}+5}} \cdot\left(e^{(3 x+2)} \cdot 3\right)\\ =\frac{3}{2} \frac{(e^(3x+2)}{\sqrt{e^{(2 x+2)}+5}}\\$$

iii)  $$y=\log \left[\tan \left(\frac{x}{2}\right)\right]\\ Diff w.r.t. \quad x \\ \frac{d y}{d x}=\frac{d}{d x} \log \left[\tan \left(\frac{x}{2}\right)\right]\\ =\frac{1}{\tan \left(\frac{x}{2}\right)} \frac{d}{d x} \tan \left(\frac{x}{2}\right)\\ =\frac{1}{\tan \left(\frac{x}{2}\right)} \cdot \sec ^{2}\left(\frac{x}{2}\right) \cdot \frac{d}{d x}\left(\frac{x}{2}\right)\\ =\frac{1}{\tan \left(\frac{x}{2}\right)} \cdot \sec ^{2}\left(\frac{x}{2}\right) \times \frac{1}{2}\\ =\frac{cos(x/2)}{2 \cdot \sin \left(\frac{x}{2}\right) }\times \frac{1}{\cos ^{2} x}\\ =\frac{1}{2 \cdot \sin \left(\frac{x}{2}\right)cos(x/2) } \qquad\because 2sinxcosx=sin2x\\ =\frac{1}{\sin x}\\ =\operatorname{cosec} x\\$$

iV)  $$y=\sqrt{\tan \sqrt{x}}\\ \\ diff w.r.t. \quad x\\ =\frac{d y}{d x}=\frac{1}{2 \sqrt{\tan \sqrt{x}}} \cdot \frac{d}{d x}(\tan \sqrt{x})\\ =\frac{1}{2 \sqrt{\tan \sqrt{x}}} \cdot \sec ^{2} \sqrt{x} \cdot \frac{d}{d x}(\sqrt{x})\\ =\frac{1}{2 \sqrt{\tan \sqrt{x}}} \cdot \sec ^{2} \sqrt{x} \times \frac{1}{2 \sqrt{x}}\\ =\frac{\sec ^{2} \sqrt{x}}{4 \sqrt{\tan \sqrt{x}}}\\$$

V) $$\quad y=cot ^{3}\left[\log \left(x^{3}\right)\right]\\ \quad y=\cot \left[\log \left(x^{3}\right)\right]^{3}\\ Diff w.r.t. ..x\\ \frac{d y}{d x}=\frac{d}{d x} \cot \left[\log x^{3}\right]^{3}\\ =3 \cdot \cot ^{2} \log x^{3} \cdot \frac{d}{d x} \cot \left(\log \left(x^{3}\right)\right)\\ =3 \cot ^{2}\left(\log x^{3}\right) \cdot\left(-\operatorname{cosec}^{2} \log \left(x^{3}\right)\right) \cdot \frac{d}{d x} \log x^{3}\\ \left.=3 \cot ^{2}[ \log \left(x^{3}\right)\right] \cdot\left(-\operatorname{cosec}^{2} \log \left(x^{3}\right)\right) \cdot \frac{1}{x^{3}} \frac{d}{d x} x^{3}\\ =3\left(cot^{2}[ \log x^{3}\right] \cdot\left(-\operatorname{cosec}^{2} \log \left(x^{3}\right)\right) \cdot \frac{1}{x^{3}} 3 x^{2}\\ =\frac{-9 \operatorname{cosec}^{2}\left[\log \left(x^{3}\right)\right] \cdot \cot ^{2}\left[\log \left(x^{3}\right)\right]}{x}\\$$

Vi) $$\quad y=5^{\sin ^{3} x+3}\\ Taking log on both side\\ \log y=\log {5}. (\sin ^{3} x+3)\\ \log y=\left(\sin ^{3} x+3\right) \log 5\\ Diff w.r.t. x\\ \frac{1}{y} \frac{d y}{d x}=\log 5 \cdot \frac{d}{d x}\left(\sin ^{3} x+3\right)\\ \qquad =\log 5\left[\frac{d}{d x} \sin ^{3} x+\frac{d}{d x} 3\right]\times y\\ \qquad =\log 5\left[3 \sin ^{2} x \cdot \frac{d}{d x} \sin x\right] \times y\\ \qquad =\log 5\left[3 \cdot \sin ^{2} x \cdot \cos x\right] \times y\\ \qquad =3 \cdot \sin ^{2} x \cdot \cos x \cdot\left(5^{\sin ^{3} x+3}\right) \cdot(\log 5)\\$$

Vii) $$y=\operatorname{cosec}(\sqrt{\cos x})\\ Diff w.r.t \quad x\\ \frac{d y}{d x}=\frac{d}{d x}(\operatorname{cosec} \sqrt{\cos x})\\ =\operatorname{cosec} \sqrt{\cos x} \cdot \cot \sqrt{\cos x} \cdot \frac{d}{d x} \sqrt{\cos x}\\ =\operatorname{cosec} \sqrt{\cos x} \cdot \cot \sqrt{\cos x} \cdot \frac{1}{2 \sqrt{\cos x}} \times \frac{d}{d x} \cos x\\ =\operatorname{cosec} \sqrt{\cos x} \cdot \cot \sqrt{\cos x} \cdot \frac{1}{2 \sqrt{\cos x}} \times(-\sin x)\\ =\frac{-\sin x \operatorname{cosec} \sqrt{\cos x} \cdot \cot \sqrt{\cos x}}{2 \sqrt{\cos x}}\\$$

Viii))$$y=\log \left[\cos \left(x^{3}-5\right)\right]\\ Diff w.r.t. x\\ \frac{d y}{d x}=\frac{d}{d x} \cdot \log \left[\cos \left(x^{3}-5\right)\right]\\ \quad=\frac{1}{\cos \left(x^{3}-5\right)} \cdot \frac{d}{d x}\left[\cos \left(x^{3}-5\right)\right]\\ \quad =\frac{1}{\cos \left(x^{3}-5\right)} \cdot\left(-\sin \left(x^{3}-5\right) \cdot \frac{d}{d x}\left(x^{3}-5\right)\right.\\ \quad =\frac{1}{\cos \left(x^{3}-5\right)}\left(-\sin \left(x^{3}-5\right)\right) \cdot\left(3 x^{2}\right)\\ \quad =-3 x^{2} \frac{\sin \left(x^{3}-5\right)}{\cos \left(x^{3}-5\right)}\\ \quad =-3 x^{2} \tan \left(x^{3}-5\right)\\$$

iX) $$y=e^{3 \sin ^{2} x}-2 \cos ^{2} x\\ Diff w.r.t. x\\ \frac{d y}{d x}=\frac{d}{d x} e^{3 \sin ^{2} x-2 \cos ^{2} x}\\ =e^{\left(3 \sin ^{2} x-2 \cos ^{2} x\right)}\frac{d}{d x}\left(3 \sin ^{2} x-2 \cos ^{2} x\right)\\ =e^{\left(3 \sin ^{2} x-2 \cos ^{2} x\right)}\left[\frac{d}{d x} 3 \sin ^{2} x-\frac{d}{d x} 2 \cos ^{2} x\right]\\ =e^{\left(3 \sin ^{2} x-2 \cos ^{2} x\right)}[3.2 \sin x \cdot \frac{d}{d x} \sin x-2 \cdot 2 \cos x \frac{d}{d x}(\cos x)]\\ =e^{\left(3 \sin ^{2} x-2 \cos ^{2} x\right)}[3 \cdot 2 \cdot \sin x \cdot \cos x+2 \cdot 2 \cdot \cos \cdot \sin x]\\ =e^{\left(3 \sin ^{2} x-2 \cos ^{2} x\right)}[3 \cdot \sin 2 x+2 \cdot \sin 2 x] \qquad [\because 2sinxcosx=sin2x]\\ =5 \sin 2 x \cdot e^{\left(3 \sin ^{2} x-2 \cos ^{2} x\right)}\\$$

X) $$ \quad y=\cos ^{2}\left[\log \left(x^{2}+7\right)\right]\\ Diff wrt \quad x\\ \frac{d y}{d x}=\frac{d}{d x} \cos \left[\log \left(x^{2}+7\right)\right]^{2}\\ =2 \cdot \cos \left[\log \left(x^{2}+7\right)\right] \cdot \frac{d}{d x} \cos \left[\log \left(x^{2}+7\right)\right]\\ =2 \cdot \cos \left[\log \left(x^{2}+7\right)\right] \cdot\left(-\sin \log \left(x^{2}+7\right)\right] \frac{d}{d x} \log \left(x^{2}+7\right)\\ =2 \cos \left[\log \left(x^{2}+7\right)\right]\left(-\sin \log \left(x^{2}+7\right)\right] \frac{1}{x^{2}+7} \cdot \frac{d}{d x}\left(x^{2}+7\right)\\ =2 \cdot \cos \left[\log \left(x^{2}+7\right)\right]\left(-\sin \left(\log \left(x^{2}+7\right) \cdot \frac{1}{x^{2}+7}\right) \cdot 2 x\right.\\ =\frac{-2 x}{x^{2}+7} \sin 2\left[\log \left(x^{2}+7\right)\right] \qquad [\because 2sinxcox=sin2x]\\$$

Xi)$$y=\tan [\cos (\sin x)]\\ Diff w.r.t.\quad x\\ \frac{d y}{d x}=\frac{d}{d x} \tan [\cos (\sin x)]\\ =\sec ^{2}[\cos (\sin x)] \cdot \frac{d}{d x}[\cos (\sin x)]\\ =\sec ^{2}[\cos (\sin x)]\left(-\sin (\sin x) \cdot \frac{d}{d x} \sin x\right.\\ =\sec ^{2}[\cos (\sin x)][-\sin (\sin x)] \cdot \cos x\\ =-\sec ^{2}[\cos (\sin x)] \cdot \sin (\sin x) \cdot \cos x\\$$

Xii)$$y=\sec \left[\tan \left(x^{4}+4\right)\right]\\ Diff w.r.t \quad x\\ \frac{d y}{d x}=\frac{d}{d x} \sec \left[\tan \left(x^{4}+4\right)\right]\\ =\sec \left(\tan \left(x^{4}+4\right)\right] \cdot \tan \left[\tan \left(x^{4}+4\right)]\right.\frac{d}{d x}\left(\tan \left(x^{4}+4\right)\right]\\ =\sec \left(\tan \left(x^{4}+4\right)\right] \cdot \tan \left[\tan \left(x^{4}+4\right)] \sec ^{2}\left(x^{4}+4\right)\right.\frac{d}{d x}\left(x^{4}+4\right)\\ =\sec \left(\tan \left(x^{4}+4\right) \cdot \tan \left[\tan \left(x^{4}+4\right)] \cdot \sec ^{2}\left(x^{4}+4\right) \cdot 4 x^{3}\right.\right.\\ =4 x^{3} \sec ^{2}\left(x^{4}+4\right) \cdot \sec \left[\tan \left(x^{4}+4\right)\right] \tan \left[\tan \left(x^{4}+4\right)\right]\\$$

Xiii)$$ y=e^{\log \left[(\log x)^{2}-\log x^{2}\right]}\\ using e^{\log x}=x\\ y=(\log x)^{2}-2 \log x\\ Diff w.r.t. \quad x \\ \frac{d y}{d x}=\frac{d}{d x}\left[(\log x)^{2}-2 \cdot \log x\right]\\ =\frac{d}{d x}(\log (x)]^{2}-2.\frac{d }{d x}\log x\\ = 2. \log x \cdot \frac{d}{d x} \log x-2 \times\frac{1}{x}\\ =\frac{2. \log x}{x}-\frac{2}{x}\\$$

Xiv)$$ y=\sin \sqrt{\sin \sqrt{x}}\\ Diff w.r.t.\quad x\\ \frac{d y}{d x}=\frac{d}{d x} \sin \sqrt{\sin \sqrt{x}}\\ =\cos \sqrt{\sin \sqrt{x}} \cdot \frac{d}{d x} \sqrt{\sin \sqrt{x}}\\ =\cos \sqrt{\sin \sqrt{x}} \cdot \frac{1}{2 \sqrt{\sin \sqrt{x}}} \cdot \frac{d}{d x}{\sin \sqrt{x}}\\ =\cos \sqrt{\sin \sqrt{x}} \cdot \frac{1}{2 \sqrt{\sin \sqrt{x}}} \cdot \cos \sqrt{x} \cdot \frac{d}{d x} \sqrt{x}\\ =\cos \sqrt{\sin \sqrt{x}} \frac{1}{2 \sqrt{\sin \sqrt{x}}} \cdot \cos \sqrt{x} \times \frac{1}{2 \sqrt{x}}\\ =\frac{\cos \sqrt{\sin \sqrt{x}} \cdot \cos \sqrt{x}}{4 \sqrt{x} \sqrt{\sin \sqrt{x}}}\\$$

Xv))$$ y=\log \left[\sec \left(e^{x^{2}}\right)\right]\\ Diff w.r.t \quad x \\ \frac{d y}{d x}=\frac{d}{d x} \log \left[\sec \left(e^{x^{2}}\right)\right]\\ =\frac{1}{\sec \left(e^{x^{2}}\right)} \cdot \frac{d}{d x} \sec \left(e^{x^{2}}\right)\\ =\frac{1}{\sec \left(e^{x^{2}}\right)} \cdot \operatorname{scc}\left(e^{x^{2}}\right) \cdot \tan \left(e^{x^{2}}\right) \cdot \frac{d}{d x} e^{x^{2}}\\ =\tan \left(e^{x^{2}}\right) \cdot e^{x^{2}} \cdot \frac{d}{d x} x^{2}\\ =\tan \left(e^{x^{2}}\right) \cdot e^{x^{2}} \cdot 2 x\\ =2 x \cdot e^{x^{2}} \cdot \tan \left(e^{x^{2}}\right)\\$$

Xvi))$$y=\log _{e} 2(\log x)\\ Diff w.r.t. \quad x\\ \frac{d y}{d x}=\frac{1}{2 \log x} \cdot \frac{d}{d x} \log x\\ =\frac{1}{2 \log x} \times \frac{1}{x}\\ =\frac{1}{2 x \cdot \log x}\\$$

Xvii)$$ y=[\log [\log (\log x)]]^2\\ Diff w.rt. \quad x \\ \frac{d y}{d x}=\frac{d}{d x}[\log [\log (\log x)]]^{2}\\ =2 \cdot \log [\log (\log x)] \cdot \frac{d}{d x} \log [\log (\log x)]\\ =2 \cdot \log (\log \log (x)] \cdot \frac{1}{\log (\log x)} \cdot \frac{d}{d x} \log [\log (x)]\\ =2 \cdot \log \left[\log (\log (x)] \frac{1}{\log (\log x)} \cdot \frac{1}{\log x} \frac{d}{d x} \log x\right.\\ =\frac{2 \cdot \log [\log (\log x)]}{\log [\log (x)] \cdot \log x} \times \frac{1}{x}\\ =\frac{2 \log (\log \log (x)]}{x \log x \log \log x)}\\$$

Xviii)$$ y=\sin ^{2} x^{2}-\cos ^{2} x^{2}\\ Diff w.r.t. \quad x\\ \frac{d y}{d x}=\frac{d}{d x}\left[\left(\sin x^{2}\right)^{2}-\left(\cos x^{2}\right)^{2}\right]\\ =\frac{d}{d x}\left(\sin x^{2}\right)^{2}-\frac{d}{d x}\left(\cos x^{2}\right)^{2}\\ =2 \cdot \sin x^{2} \cdot \frac{d}{d x} \sin x^{2}-2 \cdot \cos x^{2} \frac{d}{d x} \cos x^{2}\\ =2 \cdot \sin x^{2} \cdot \cos x^{2} \cdot \frac{d}{d x} x^{2}+2 \cdot \cos x^{2} \cdot \sin ^{2} x^{2} \frac{d}{d x} x^{2}\\ =\sin 2 x^{2} \cdot 2 x+ \sin 2 x^{2} \cdot 2 x \quad \because 2sinxcox=sin2x\\ =\quad 4 x \cdot \sin \left(2 x^{2}\right)$$

• Q.1 --Link
• Q.2 -- Link
• Q.3 -- Link
• Q.4, Q.5, Q.6 --Link
• Q.7 -- Link

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Maharashtra State Board HSC 12th Mathematics Book exercise 1.1 differentiation Solutions 2021 new
Maharashtra State Board HSC  12th Mathematics Book exercise 1.1 differentiation Solutions 2021 new Solutions
Maharashtra State Board HSC  12th Mathematics textbook exercise 1.1 differentiation Solutions 2021 new
Maharashtra State Board HSC  12th Mathematics textbook exercise 1.1 differentiation Solutions 2021 new