Question

pls help me out with my physics prblms​

Answer 1

$\underline{\underline{\sf \qquad Solution : \qquad}}$

$\longrightarrow\:\:\sf y = x^{-3}$

On differentiating both the sides with respect to x, we get :

$\longrightarrow\:\:\sf \dfrac{d}{dx} {x}^{n} =n {x}^{n - 1}$

$\longrightarrow\:\:\sf \dfrac{dy}{dx} = - 3 {x}^{ - 3 - 1}$

$\longrightarrow\:\:\sf \dfrac{dy}{dx} = - 3 {x}^{ - 4}$

$\longrightarrow\: \: \underline{ \boxed{\sf \dfrac{dy}{dx} = \dfrac{ - 3}{ {x}^{4} } }}$

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$\longrightarrow\:\:\sf 6x^{5} + 4x^{3} - 3x^{2} + 2x - 7$

• Let y = 6x⁵ + 4x³- 3x² + 2x - 7

$\longrightarrow\:\:\sf \dfrac{d}{dx} {x}^{n} =n {x}^{n - 1} \: and \: \dfrac{d}{dx} c=0$

On differentiating both the sides with respect to x, we get :

$\longrightarrow\:\:\sf \dfrac{dy}{dx} = 6.5x^{5 - 1} + 4.3x^{3 - 1} - 3.2x^{2-1} + 2.x^{1 - 1} - 0$

$\longrightarrow\:\:\sf \dfrac{dy}{dx} = 6.5x^{4} + 4.3x^{2} - 3.2x^{1} + 2 \times 1$

$\longrightarrow\:\: \underline{ \boxed{\sf \dfrac{dy}{dx} = 30x^{4} + 12x^{2} - 6x + 2 }}$

Answer 2

$\text{(I)} y = x^{-3}$

$\text{It's given }y = x^{-3}$

$\frac{d}{dx}(x^n)= nx^{n-1}$

$\frac{d}{dx}(x^{-3})=(-3) x^{-3-1}$

$\implies -3x^{-4}$      $(\therefore a^{-b} = \frac{1}{a^b})$

$\boxed{\frac{dy}{dx} = \frac{-3}{x^4}}$

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$\text{(II)}6x^5+4x^3-3x^2+2x-7$

$\text{Let } y = 6x^5+4x^3-3x^2+2x-7$

$\frac{d}{dx}(6x^5) \implies \frac{d}{dx}(x^n)\implies nx^{n-1} \implies (5)6x^{5-1}$

$\frac{d}{dx}(4x^3) \implies \frac{d}{dx}(x^n)\implies nx^{n-1} \implies (3)4x^{3-1}$

$\frac{d}{dx}(-3x^2) \implies \frac{d}{dx}(x^n) \implies nx^{n-1}\implies (2)-3x^{2-1}$

$\frac{d}{dx}(2x) \implies \frac{d}{dx}(kx) \implies k \implies 2$

$\frac{d}{dx}(-7)\implies\frac{d}{dx}(k)\implies 0$

$\text{Now substitute the differentiated values}$

$= (5)6x^{5-1} + (3)4x^{3-1} - (2)3x^{2-1}+2 -0$

$= (5)6x^4 + (3)4x^2-(2)3x +2$

$\boxed{\frac{dy}{dx} = 30x^4+12x^2- 6x+2}$