Question

Please solve these all questions in less explanation. (6 Questions)

Answer 1

$\large\underline{\sf{Solution-32}}$

$\rm \: y = {\bigg( {2x}^{4} + {3x}^{3} - 5x + 6\bigg)}^{ - \dfrac{1}{3} }$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{dy}{dx} = - \frac{1}{3} {\bigg( {2x}^{4} + {3x}^{3} - 5x + 6\bigg)}^{ - \dfrac{1}{3} - 1}\dfrac{d}{dx}({2x}^{4} + {3x}^{3} - 5x + 6)$

$\rm \: \dfrac{dy}{dx} = - \frac{1}{3} {\bigg( {2x}^{4} + {3x}^{3} - 5x + 6\bigg)}^{ - \dfrac{4}{3}}({8x}^{3} + {9x}^{2} - 5)$

So, option (a) is correct.

$\large\underline{\sf{Solution-33}}$

$\rm \: y = log[ \sqrt{x - 1} - \sqrt{x + 1}]$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{dy}{dx} = \frac{1}{ \sqrt{x - 1} - \sqrt{x + 1} }\dfrac{d}{dx} [ \sqrt{x - 1} - \sqrt{x + 1}]$

$\rm \: \dfrac{dy}{dx} = \frac{1}{ \sqrt{x - 1} - \sqrt{x + 1} }\bigg(\dfrac{1}{2 \sqrt{x - 1} } - \dfrac{1}{2 \sqrt{x + 1} } \bigg)$

$\rm \: \dfrac{dy}{dx} = \frac{1}{ \sqrt{x - 1} - \sqrt{x + 1} }\bigg( \dfrac{ \sqrt{x + 1} - \sqrt{x - 1} }{2 \sqrt{x + 1} \sqrt{x + 1} } \bigg)$

$\rm \: \dfrac{dy}{dx} = - \frac{1}{ \sqrt{x - 1} - \sqrt{x + 1} }\bigg( \dfrac{ \sqrt{x - 1} - \sqrt{x + 1} }{2 \sqrt{ {x}^{2} - 1} } \bigg)$

$\rm \: \dfrac{dy}{dx} = - \dfrac{ 1 }{2 \sqrt{ {x}^{2} - 1} }$

$\rm \: \dfrac{dy}{dx} = - \dfrac{ 1 }{2} {\bigg( {x}^{2} - 1 \bigg) }^{ - \dfrac{1}{2} }$

So, option (b) is correct.

$\large\underline{\sf{Solution-34}}$

$\rm \: y = log \sqrt{x + \sqrt{ {x}^{2} + {a}^{2} } }$

can be rewritten as

$\rm \: y = \frac{1}{2} \: log[x + \sqrt{ {x}^{2} + {a}^{2}}]$

On differentiating both sides w. r. t. x,

$\rm \: \dfrac{dy}{dx} = \frac{1}{2(x + \sqrt{ {x}^{2} + {a}^{2} }) }\dfrac{d}{dx}[x + \sqrt{ {x}^{2} + {a}^{2}}]$

$\rm \: \dfrac{dy}{dx} = \frac{1}{2(x + \sqrt{ {x}^{2} + {a}^{2} }) }\bigg[1 + \frac{1}{2\sqrt{ {x}^{2} + {a}^{2}}} \times 2x\bigg]$

$\rm \: \dfrac{dy}{dx} = \frac{1}{2(x + \sqrt{ {x}^{2} + {a}^{2} }) }\bigg[1 + \frac{x}{\sqrt{ {x}^{2} + {a}^{2}}}\bigg]$

$\rm \: \dfrac{dy}{dx} = \frac{1}{2(x + \sqrt{ {x}^{2} + {a}^{2} }) }\bigg[ \frac{\sqrt{ {x}^{2} + {a}^{2}} + x}{\sqrt{ {x}^{2} + {a}^{2}}}\bigg]$

$\rm \: \dfrac{dy}{dx} = \frac{1}{2\sqrt{ {x}^{2} + {a}^{2}}}$

$\rm \: \dfrac{dy}{dx} = \dfrac{ 1 }{2} {\bigg( {x}^{2} + {a}^{2} \bigg) }^{ - \dfrac{1}{2} }$

So, option (a) is correct

$\large\underline{\sf{Solution-35}}$

$\rm \: x = \frac{3at}{1 + {t}^{3} }$

$\rm \: \dfrac{dx}{dt} = 3a \: \frac{(1 + {t}^{3}).1 - t( {3t}^{2} )}{(1 + {t}^{3})^{2} }$

$\rm \: \dfrac{dx}{dt} = 3a \: \frac{1 + {t}^{3} - {3t}^{3}}{(1 + {t}^{3})^{2} }$

$\rm \: \dfrac{dx}{dt} = \: \frac{3a(1 - {2t}^{3})}{(1 + {t}^{3})^{2} }$

Now,

$\rm \: y = \frac{3a {t}^{2} }{1 + {t}^{3} }$

$\rm \: \dfrac{dy}{dt} = 3a \: \frac{(1 + {t}^{3}). {2t} - {t}^{2} ( {3t}^{2} )}{(1 + {t}^{3})^{2} }$

$\rm \: \dfrac{dy}{dt} = 3a \: \frac{2t + 2{t}^{4} - 3 {t}^{4}}{(1 + {t}^{3})^{2} }$

$\rm \: \dfrac{dy}{dt} = \: \frac{3a(2t - {t}^{4})}{(1 + {t}^{3})^{2} }$

So,

$\rm \: \dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}$

$\rm \: \dfrac{dy}{dx} = \frac{2t - {t}^{4} }{1 - {2t}^{3} }$

So, option (a) is correct.

$\large\underline{\sf{Solution-36}}$

$\rm \: y = log\bigg( {e}^{3x} {\bigg(5x - 3 \bigg) }^{\dfrac{1}{3} } {\bigg(4x + 2\bigg) }^{ - \dfrac{1}{3} } \bigg)$

$\rm \: y = 3x + \dfrac{1}{3}log(5x - 3) - \dfrac{1}{3}log(4x + 2)$

$\rm \: \dfrac{dy}{dx} = 3 + \dfrac{1}{3(5x - 3)} \times 5 - \dfrac{1}{3(4x + 2)} \times 4$

$\rm \: \dfrac{dy}{dx} = 3 + \frac{1}{3} \bigg(\dfrac{5}{(5x - 3)} - \dfrac{4}{(4x + 2)}\bigg)$

So, option (a) is correct.

$\large\underline{\sf{Solution-38}}$

$\rm \: {x}^{y} = {e}^{x - y}$

$\rm \: log{x}^{y} = log {e}^{x - y}$

$\rm \: ylogx = x - y$

$\rm \: ylogx + y = x$

$\rm \: y(logx + 1) = x$

$\rm \: y = \frac{x}{1 + logx}$

$\rm \: \dfrac{dy}{dx} = \frac{(logx + 1).1 - x \times \dfrac{1}{x} }{(1 + logx) ^{2} }$

$\rm \: \dfrac{dy}{dx} = \frac{logx + 1 - 1 }{(1 + logx) ^{2} }$

$\rm \: \dfrac{dy}{dx} = \frac{logx }{(1 + logx) ^{2} }$

So, option (b) is correct

Answer 2

Answer:

32.(a)

33.(b)

34.(a)

35.(a)

36.(a)

37.(c)

38.(b)

Step-by-step explanation:

Hope it helps