Differentiation Exercise 1.1
Differentiation Exercise 1.1 12th Class Math Maharashtra Board
2 Differentiate the following w.r.t x
$(i)cos(x^2+a^2)$
Solution-
$y=cos(x^2+a^2)$
differentiate w.r.t x
$\frac{dy}{dx}=-sin(x^2+a^2)$$\frac{d}{dx}(x^2+a^2)$
$(ii)\sqrt{e^{3x+2}+5}$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{1}{\sqrt{e^{3x+2}+5}}\frac{d}{dx}(e^{3x+2}+5)$
$=\frac{1}{2\sqrt{e^{3x+2}+5}}e^{3x+2}\frac{d}{dx}(3x+2)+0 $
$=\frac{1}{2\sqrt{e^{3x+2}+5}}e^{3x+2}(3\times 1+0) $
$=\frac{3}{2\sqrt{e^{3x+2}+5}}e^{3x+2} $
$(iii)\sqrt{tan\sqrt{tan(x)}}$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{1}{2\sqrt{tan\sqrt{tan(x)}}}\times\frac{d}{dx}tan\sqrt{tan(x)}$
$=\frac{1}{2\sqrt{tan\sqrt{tan(x)}}}\times sec^2(\sqrt{tan(x)})\frac{d}{dx}\sqrt{tan(x)}$
$=\frac{1}{2\sqrt{tan\sqrt{tan(x)}}}\times sec^2(\sqrt{tan(x)})\frac{1}{2\sqrt{tan(x)}}\frac{d}{dx}tan(x)$
$\frac{dy}{dx}=\frac{sec^2(\sqrt{tan(x)})}{2\sqrt{tan\sqrt{tan(x)}}}\times \frac{sec^2x}{2\sqrt{tan(x)}}
$(iv)\sqrt{tan\sqrt{x}}$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}\sqrt{tan\sqrt{x}}$
$=\frac{1}{2\sqrt{tan\sqrt{x}}}\frac{d}{dx}tan\sqrt{x}$
$=\frac{1}{2\sqrt{tan\sqrt{x}}}sec^2\sqrt{x}\frac{d}{dx}\sqrt x$
$=\frac{1}{2\sqrt{tan\sqrt{x}}}sec^2\sqrt{x}\frac{1}{2\sqrt x}$
$=\frac{sec^2\sqrt{x}}{4\sqrt x\sqrt{tan\sqrt{x}}}$
$(v)cot^3\left[log(x^3)\right]$
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}cot^3\left[log(x^3)\right]$
$=3cot^2\left[log(x^3)\right]\frac{d}{dx}\left[log(x^3)\right]$
$=3cot^2\left[log(x^3)\right]\frac{1}{\left[log(x^3)\right]}\frac{d}{dx}(x^3)$
$=3cot^2\left[log(x^3)\right]\frac{3x^2}{\left[log(x^3)\right]}$
$(vi)5^{sin^3x+3}$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}(5^{sin^3x+3})$
$=5^{sin^3x+3}\times log 5\times\frac{d}{dx}(sin^3x+3)$
$=5^{sin^3x+3}\times log 5\times(3sin^2x\frac{d}{dx}(sinx)+0)$
$\frac{dy}{dx}=5^{sin^3x+3}\times log 5\times(3sin^2xcosx)$
$(vii)cosec(\sqrt cos x)$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}cosec(\sqrt{cos x})$
$=-cosec(\sqrt{cos x})sec(\sqrt{cos x})\frac{d}{dx}(\sqrt{cos x})$
$=-cosec(\sqrt{cos x})sec(\sqrt{cos x})\frac{1}{2\sqrt {cos x}}\frac{d}{dx} cos x$
$=-cosec(\sqrt{cos x})sec(\sqrt{cos x})\frac{-sin x}{2\sqrt {cos x}}$
$=\frac{sin xcosec(\sqrt{cos x})sec(\sqrt{cos x})}{2\sqrt {cos x}}$
$(viii)log[cos(x^3-5)$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}(log[cos(x^3-5)])$
$=\frac{1}{cos(x^3-5)}\frac{d}{dx}[cos(x^3-5)]$
$=\frac{1}{cos(x^3-5)}[-sin(x^3-5)\frac{d}{dx}(x^3)-0]$
$=\frac{-3x^2sin(x^3-5)}{cos(x^3-5)}$
$=-3x^2tan(x^3-5)$
$(ix)e^{3 sin^2x-2cos^2x}$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}(e^{3 sin^2x-2cos^2x})$
$=e^{3 sin^2x-2cos^2x}\frac{d}{dx}(3 sin^2x-2cos^2x)$
$=e^{3 sin^2x-2cos^2x}\times (6sin x\frac{d}{dx}sinx -4cos x\frac{d}{dx}cos x)$
$=e^{3 sin^2x-2cos^2x}\times(6sinx.cosx+4cosx.sinx)$
$(x)cos^2\left[log(x^2+7)\right]$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}cos^2\left[log(x^2+7)\right]$
$=2cos\left[log(x^2+7)\right]\frac{d}{dx}cos\left[log(x^2+7)\right]$
$=-2cos\left[log(x^2+7)\right]sin\left[log(x^2+7)\right]\frac{d}{dx}\left[log(x^2+7)\right]$
$=-2cos\left[log(x^2+7)\right]sin\left[log(x^2+7)\right]\frac{1}{(x^2+7)}\frac{d}{dx}(x^2+7)$
$=-2cos\left[log(x^2+7)\right]sin\left[log(x^2+7)\right]\frac{1}{(x^2+7)}(2x+0)$
$=-2cos\left[log(x^2+7)\right]\frac{2xsin\left[log(x^2+7)\right]}{(x^2+7)}$
$=\frac{-2xsin\left[log(x^2+7)\right]}{x^2+7}$
$(xi)tan\left[cos(sin x)\right]$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}tan\left[cos(sin x)\right]$
$=sec^2\left[cos(sin x)\right]\frac{d}{dx}\left[cos(sin x)\right]$
$=sec^2\left[cos(sin x)\right]\times-sin(sin x)\frac{d}{dx}(sin x)$
$=-sec^2\left[cos(sin x)\right]\times sin(sin x)\times cosx $
$(xii)sec\left[x^4+4\right]$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}\{sec\left[tan(x^4+4)\right]\}$
$=sec\left[tan(x^4+4)\right]tan\left[tan(x^4+4)\right]\frac{d}{dx}\left[tan(x^4+4)\right]$
$=sec\left[tan(x^4+4)\right]tan\left[tan(x^4+4)\right]sec^2(x^4+4)\frac{d}{dx}(x^4+4)$
$=sec\left[tan(x^4+4)\right]tan\left[tan(x^4+4)\right]sec^2(x^4+4)\times(4x^3+0)$
$=4x^3.sec^2(x^4+4)sec\left[tan(x^4+4)\right]tan\left[tan(x^4+4)\right]$
$(xiii)e^{\left[(log x)^2-log x^2\right]}$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}e^{\left[(log x)^2-log x^2\right]}$
$=e^{\left[(log x)^2-log x^2\right]}\frac{d}{dx}\left[(log x)^2-log x^2\right]$
$=e^{\left[(log x)^2-log x^2\right]}\left[2(log x)\frac{d}{dx}(log x)-\frac{1}{x^2}\frac{d}{dx}x^2\right]$
$=e^{\left[(log x)^2-log x^2\right]}\left[2(log x)\frac{1}{x}-\frac{1}{x^2}2x\right]$
$=e^{\left[(log x)^2-log x^2\right]}\left[\frac{2(log x)}{x}-\frac{2}{x}\right]$
$(xiv)log_{e^2}(log x)$
Solution-
differentiate w.r.t x\\
\frac{dy}{dx}=\frac{d}{dx}sin\sqrt{sin\sqrt{x}}$
$=sin\sqrt{sin\sqrt{x}}\frac{d}{dx}\sqrt{sin\sqrt{x}}$
$=sin\sqrt{sin\sqrt{x}}\frac{1}{2\sqrt{sin\sqrt{x}}}\frac{d}{dx}sin\sqrt{x}$
$=\frac{sin\sqrt{sin\sqrt{x}}}{2\sqrt{sin\sqrt{x}}}cos\sqrt{x}\frac{d}{dx}\sqrt{x}$
$=\frac{sin\sqrt{sin\sqrt{x}cos\sqrt{x}}}{2\sqrt{sin\sqrt{x}}}\frac{1}{2\sqrt{x}}$
$=\frac{sin\sqrt{sin\sqrt{x}cos\sqrt{x}}}{4\sqrt{x}\sqrt{sin\sqrt{x}}}$
$(xv)$log[sec(e^{x}^2)]$
Solution-
differentiate w.r.t x
$(xvi)$log_{e^2}(log x)$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}log_{e^2}(log x)$
$=\frac{log (log x)}{log e^2}$
$=\frac{1}{log e^2}\frac{1}{log x}\frac{d}{dx}(log x)$
$=\frac{1}{logx loge^2}\frac{1}{x}$
$=\frac{1}{xlogx loge^2}$
$=\frac{1}{2xlogx}$
$(xvii)\left[log[log(x)]\right]$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}\left[log\left[log(log(x))\right]\right]^2$
$=2\left[log\left[log(log(x))\right]\right]\frac{d}{dx}\left[log\left[log(log(x))\right]\right]$
$=2\left[log\left[log(log(x))\right]\right]\frac{1}{log(log(x))}\frac{d}{dx}\left[log(log(x))\right]$
$=2\left[log\left[log(log(x))\right]\right]\frac{1}{log(log(x))}\frac{1}{log x}\frac{d}{dx}(log x)$
$=2\left[log\left[log(log(x))\right]\right]\frac{1}{log(log(x))}\frac{1}{log x}\frac{1}{x}$
$=\frac{2\left[log\left[log(log(x))\right]\right]}{xlog(log x)log x}$
$(xviii)sin^2x^2-cos^2x^2$
Solution-
differentiate w.r.t x
$\frac{dy}{dx}=\frac{d}{dx}sin^2x^2-cos^2x^2$
$=2sinx^2\frac{d}{dx}sinx^2-2cosx^2\frac{d}{dx}cosx^2$
$=2sinx^2cosx^2\frac{d}{dx}x^2+2cosx^2sinx^2\frac{d}{dx}x^2$
$=4xsinx^2cosx^2+4xcosx^2sinx^2$
$=4x(sinx^2cosx^2+cosx^2sinx^2)$
$=4xsin(2x^2)$
Differentiation Exercise 1.1 question 2) pdf
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Differentiation Exercise 1.1 |