Question

y=1/2x+3 then dy/dx at x=0 is |2/n| find the value of n

Answer 1

Answer:4/ln(3)

Explanation:If Y =1/2x +3 then dy/dx = 1/2 ln(2x +3) , so 1/2 ln (2x+3) = 2/n . At x =0 , 1/2 ln(3) = 2/n ,n = 4 /ln(3)

Answer 2

Answer:

Required value of n is -9.

Explanation:

Given

$y = \frac{1}{2x + 3}$

$y = {(2x + 3)}^{ - 1}$

we are differentiating y with respect to x,

$\frac{dy}{dx} = \frac{d}{dx} {(2x + 3)}^{ - 1}$

$\frac{dy}{dx} =( - 1) {(2x + 3)}^{ - 2} \frac{d}{dx} (2x + 3)$

$\frac{dy}{dx} = - {(2x + 3)}^{ - 2} \times 2$

Here we are putting x=0,

Then $\frac{dy}{dx}$$= - {(2 \times 0 + 3)}^{ - 2} \times 2 = - 2 \times {3}^{ - 2} = - 2 \times \frac{1}{9} = - \frac{2}{9}$

Given when x= 0,

$\frac{dy}{dx} = | \frac{2}{n} |$

According to question,

$| \frac{2}{n} | = - \frac{2}{9}$

$\frac{2}{n} = - \frac{2}{9}$

By cross multiplication,

$2 \times 9 = - 2 \times n$

$n = - \frac{2 \times 9}{2} = - 9$

Required value of n is -9.

Here applied formulas are$\frac{d}{dx} ( {x}^{n} ) = n {x}^{n - 1}$

$\frac{d}{dx} {(ax)}^{n} = n {(ax)}^{ n - 1} \frac{d}{dx} (ax) = n {(ax)}^{ n - 1} \times a$

Some other formulas of derivative,

$1. \frac{d}{dx} ( \sin(x) ) = \cos(x ) \\ 2. \frac{d}{dx} ( \cos(x) ) = - \sin(x) \\ 3. \frac{d}{dx} ( \tan(x) ) = {sec}^{2} x \\ 4. \frac{d}{dx} ( \cot(x) ) = - {cosec}^{2} x \\ 5. \frac{d}{dx} ( \sec(x) ) = \sec(x) \tan(x) \\ 6. \frac{d}{dx} (cosec(x)) = - cosecxcotx$