Question

√3x-5-1/√3x-5 , Find derivative.

Answer 1

Let, $y=\sqrt{3x-5}-\frac{1}{\sqrt{3x-5}}$

$\Rightarrow y=\frac{3x-5-1}{\sqrt{3x-5}}$

$\Rightarrow y=\frac{3x-6}{\sqrt{3x-5}}$

Now, differentiating both sides with respect to x, we get

$\quad\frac{dy}{dx}=\frac{d}{dx}\big(\frac{3x-6}{\sqrt{3x-5}}\big)$

$\Rightarrow\frac{dy}{dx}=\frac{\sqrt{3x-5}\frac{d}{dx}(3x-6)-(3x-6)\frac{d}{dx}(\sqrt{3x-5})}{(\sqrt{3x-5})^{2}}$

$\Rightarrow\frac{dy}{dx}=\frac{3\sqrt{3x-5}-\frac{3(3x-6)}{2\sqrt{3x-5}}}{3x-5}$

$\Rightarrow\frac{dy}{dx}=\frac{3\sqrt{3x-5}-\frac{9(x-2)}{2\sqrt{3x-5}}}{3x-5}$

$\Rightarrow\frac{dy}{dx}=\frac{\frac{6(3x-5)-9(x-2)}{\sqrt{3x-5}}}{3x-5}$

$\Rightarrow\frac{dy}{dx}=\frac{18x-30-9x+18}{(3x-5)\sqrt{3x-5}}$

$\Rightarrow\frac{d}{dx}\big(\sqrt{3x-5}-\frac{1}{\sqrt{3x-5}}\big)=\frac{9x-12}{(3x-5)\sqrt{3x-5}}$

This is the required derivative.

Answer 2

dy/dx = √(3x - 5) - 1/√(3x - 5) = (9x - 12)/(2∛(3x - 5))

Step-by-step explanation:

√(3x - 5) - 1/√(3x - 5) = (3x - 6)/√(3x - 5)

On applying quotient rule, we get,

d(f(x)/g(x)) = ((g(x) × df(x)) - (f(x) - dg(x)))/(g(x))²

Now,

y = f(x)/g(x) = (3x - 6)/√(3x - 5)

df(x) = 3

dg(x) = 3/(2(3x - 5))¹⁾²

Thus,

d(f(x)/g(x)) = ((√(3x - 5) × 3) - ((3x - 6) × 3/(2√(3x - 5))))/(3x - 5)

d(f(x)/g(x)) = (3(√(3x - 5)) - ((3(3x - 6))/(2√(3x - 5)))/(3x - 5)

d(f(x)/g(x)) = (6(3x - 5) - 3(3x - 6))/(2√(3x - 5))/(3x - 5)

d(f(x)/g(x)) = (6(3x - 5) - 3(3x - 6))/(2√(3x - 5)(3x - 5))

d(f(x)/g(x)) = (18x - 30 - 9x + 18)/(2∛(3x - 5))

d(f(x)/g(x)) = (9x - 12)/(2∛(3x - 5))