Class 12 Maths NCERT Solutions Chapter 5 Miscellaneous Exercise

u=3x^2-9x+5,\quad u'=6x-9

\dfrac{dy}{dx}=9u^8\cdot u'=9(6x-9)(3x^2-9x+5)^8

\dfrac{dy}{dx}=3\sin^2x\cos x+6\cos^5x\cdot(-\sin x)=3\sin^2x\cos x-6\sin x\cos^5x

=3\sin x\cos x\,(\sin x-2\cos^4x)

Let y=(5x)^{3\cos 2x}. Taking log: \log y = 3\cos 2x\,\log(5x)

\dfrac{1}{y}\dfrac{dy}{dx}=-6\sin 2x\,\log(5x)+\dfrac{3\cos 2x}{x}

x\sqrt{x}=x^{3/2}

\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^3}}\cdot\dfrac{3}{2}x^{1/2}

u=\cos^{-1}\dfrac{x}{2},\qquad v=(2x+7)^{1/2}

u'=\dfrac{-1/2}{\sqrt{1-x^2/4}}=\dfrac{-1}{\sqrt{4-x^2}},\qquad v'=\dfrac{1}{\sqrt{2x+7}}

\dfrac{dy}{dx}=\dfrac{u'v-uv'}{v^2}=\dfrac{\dfrac{-\sqrt{2x+7}}{\sqrt{4-x^2}}-\dfrac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}}}{2x+7}

Combine over the common denominator (2x+7)^{3/2}\sqrt{4-x^2}:

1+\sin x=\left(\sin\tfrac{x}{2}+\cos\tfrac{x}{2}\right)^2,\qquad 1-\sin x=\left(\cos\tfrac{x}{2}-\sin\tfrac{x}{2}\right)^2

For 0<x<\tfrac{\pi}{2}: numerator =2\cos\tfrac{x}{2}, denominator =2\sin\tfrac{x}{2}

y=\cot^{-1}\left(\cot\tfrac{x}{2}\right)=\tfrac{x}{2}\;\Rightarrow\;\dfrac{dy}{dx}=\dfrac{1}{2}

Let y=(\log x)^{\log x}. Taking log: \log y=\log x\cdot\log(\log x)

\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{x}\log(\log x)+\dfrac{1}{x}=\dfrac{1}{x}\big[1+\log(\log x)\big]

\dfrac{dy}{dx}=-\sin(a\cos x+b\sin x)\cdot(-a\sin x+b\cos x)

Let t=\sin x-\cos x. \log y=t\log t

\dfrac{1}{y}\dfrac{dy}{dx}=(\cos x+\sin x)\big[1+\log t\big]

\dfrac{d}{dx}(x^x)=x^x(1+\log x) (log differentiation)

\dfrac{d}{dx}(x^a)=ax^{a-1},\qquad \dfrac{d}{dx}(a^x)=a^x\log a,\qquad \dfrac{d}{dx}(a^a)=0

u=x^{x^2-3}:\quad \log u=(x^2-3)\log x\;\Rightarrow\;u'=x^{x^2-3}\left[2x\log x+\dfrac{x^2-3}{x}\right]

v=(x-3)^{x^2}:\quad \log v=x^2\log(x-3)\;\Rightarrow\;v'=(x-3)^{x^2}\left[2x\log(x-3)+\dfrac{x^2}{x-3}\right]

\dfrac{dy}{dt}=12\sin t,\qquad \dfrac{dx}{dt}=10(1-\cos t)

\dfrac{dy}{dx}=\dfrac{12\sin t}{10(1-\cos t)}=\dfrac{6}{5}\cdot\dfrac{2\sin\frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}}=\dfrac{6}{5}\cot\dfrac{t}{2}

\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^2}}

With u=\sqrt{1-x^2}: 1-u^2=x^2\;\Rightarrow\;\sqrt{1-u^2}=x (since 0<x<1); u'=\dfrac{-x}{\sqrt{1-x^2}}

\dfrac{d}{dx}\sin^{-1}u=\dfrac{1}{x}\cdot\dfrac{-x}{\sqrt{1-x^2}}=\dfrac{-1}{\sqrt{1-x^2}}

\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}-\dfrac{1}{\sqrt{1-x^2}}=0

Given: x\sqrt{1+y}+y\sqrt{1+x}=0  …(1)

Rearrange (1): x\sqrt{1+y}=-y\sqrt{1+x}

Square both sides: x^2(1+y)=y^2(1+x)

(x^2-y^2)+(x^2y-xy^2)=0\;\Rightarrow\;(x-y)(x+y)+xy(x-y)=0\;\Rightarrow\;(x-y)(x+y+xy)=0

Since x\ne y: x+y+xy=0\;\Rightarrow\;y=\dfrac{-x}{1+x}

\dfrac{dy}{dx}=\dfrac{-(1+x)+x}{(1+x)^2}=\dfrac{-1}{(1+x)^2}

Given: (x-a)^2+(y-b)^2=c^2  …(1)

Differentiate (1) w.r.t. x: 2(x-a)+2(y-b)y'=0\;\Rightarrow\;(x-a)=-(y-b)y'  …(2)

Differentiate (2) w.r.t. x (product rule on the right side): 1+(y')^2+(y-b)y''=0\;\Rightarrow\;(y-b)=\dfrac{-\big[1+(y')^2\big]}{y''}  …(3)

Square (2): (x-a)^2=(y-b)^2(y')^2

Substitute into (1): (y-b)^2(y')^2+(y-b)^2=c^2\;\Rightarrow\;(y-b)^2\big[1+(y')^2\big]=c^2  …(4)

Square (3): (y-b)^2=\dfrac{\big[1+(y')^2\big]^2}{(y'')^2}

Substitute into (4): \dfrac{\big[1+(y')^2\big]^2}{(y'')^2}\big[1+(y')^2\big]=c^2\;\Rightarrow\;\big[1+(y')^2\big]^3=c^2(y'')^2

Given: \cos y=x\cos(a+y)  …(1)

From (1): x=\dfrac{\cos y}{\cos(a+y)}  …(2)

Differentiate (1) implicitly (product rule on the right side): -\sin y\cdot y'=\cos(a+y)-x\sin(a+y)\cdot y'

y'\big[x\sin(a+y)-\sin y\big]=\cos(a+y)

Substitute (2): x\sin(a+y)-\sin y=\dfrac{\cos y\sin(a+y)-\sin y\cos(a+y)}{\cos(a+y)}=\dfrac{\sin a}{\cos(a+y)}

\dfrac{dy}{dx}=\dfrac{\cos(a+y)}{\sin a/\cos(a+y)}=\dfrac{\cos^2(a+y)}{\sin a}

\dfrac{dx}{dt}=a\big[-\sin t+\sin t+t\cos t\big]=at\cos t

\dfrac{dy}{dt}=a\big[\cos t-\cos t+t\sin t\big]=at\sin t

\dfrac{dy}{dx}=\tan t

\dfrac{d}{dt}(\tan t)=\sec^2t\;\Rightarrow\;\dfrac{d^2y}{dx^2}=\dfrac{\sec^2t}{at\cos t}=\dfrac{1}{at\cos^3t}

For x\ge0: f(x)=x^3\Rightarrow f'(x)=3x^2. For x<0: f(x)=-x^3\Rightarrow f'(x)=-3x^2

At x=0: f'(0)=\lim\limits_{h\to0}\dfrac{|h|^3}{h}=0 (both sides)  \Rightarrow\;f'(x)=3x|x|

Differentiate again: for x\ge0, f''(x)=6x; for x<0, f''(x)=-6x

At x=0: f''(0)=\lim\limits_{h\to0}\dfrac{3h|h|}{h}=0

Differentiate both sides w.r.t. A (B constant): \dfrac{d}{dA}\sin(A+B)=\cos(A+B)

\dfrac{d}{dA}\big[\sin A\cos B+\cos A\sin B\big]=\cos A\cos B-\sin A\sin B

Yes: f(x)=|x|+|x-1| — continuous everywhere (sum of continuous functions).

x<0:\; f(x)=1-2x,\; f'=-2

0<x<1:\; f(x)=1,\; f'=0

x>1:\; f(x)=2x-1,\; f'=2

At x=0: LHD =-2, RHD =0 (differ). At x=1: LHD =0, RHD =2 (differ).

Expand along the first row (l, m, n, a, b, c constant): y=f(x)(mc-nb)-g(x)(lc-na)+h(x)(lb-ma)

\dfrac{dy}{dx}=f'(x)(mc-nb)-g'(x)(lc-na)+h'(x)(lb-ma)

Given: y=e^{a\cos^{-1}x}  …(1)

Differentiate (1): y'=e^{a\cos^{-1}x}\cdot a\cdot\dfrac{-1}{\sqrt{1-x^2}}=\dfrac{-ay}{\sqrt{1-x^2}}  …(2)

Move the denominator of (2) to the left, then square both sides: \sqrt{1-x^2}\,y'=-ay\;\Rightarrow\;(1-x^2)(y')^2=a^2y^2  …(3)

Differentiate (3) w.r.t. x (product rule on the left side): -2x(y')^2+(1-x^2)\cdot2y'y''=2a^2yy'

Divide throughout by 2y': -xy'+(1-x^2)y''=a^2y