Question

y=1nx/x^2+5x.differencate​

Answer 1

Answer:

(x + 5) - (lnx)(2x + 5)/(x² + 5x)²

Explanation:

As per the information provided in the question, We have

• y = lnx/x² + 5x

We are asked to find dy/dx.

In order to find dy/dx, We will Differentiate using qoutient rule.

It is given by –

${\longmapsto \rm \dfrac{d}{dx}(y) = \dfrac{d}{dx} \bigg( \dfrac{u}{v}\bigg) = \dfrac{ v \dfrac{d}{dx} (u) - u \dfrac{d}{dx} (v)}{ {(v)}^{2} }}$

Where,

• u = lnx
• v = x² + 5x

Substituting the values.

${ \longmapsto \rm \dfrac{d}{dx}(y) = \dfrac{ ( {x}^{2} + 5x) \dfrac{d}{dx} (lnx) - (lnx) \dfrac{d}{dx} ({x}^{2} + 5x)}{ {( {x}^{2} + 5x )}^{2} }}$

Differentiating lnx,

${\longmapsto \rm \dfrac{d}{dx}(y) = \dfrac{ ( {x}^{2} + 5x) \bigg( \dfrac{1}{x} \bigg) - (lnx) \dfrac{d}{dx} ({x}^{2} + 5x)}{ {( {x}^{2} + 5x )}^{2} }}$

Differentiating x² + 5x,

Using,

• ${\longmapsto \rm \dfrac{d}{dx} \times {x}^{n} = n \times {x}^{n - 1} }$

• ${\longmapsto \rm \dfrac{d}{dx} ( {x}) = 1}$

Thus, It will be,

${ \longmapsto \rm \dfrac{d}{dx}(y) = \dfrac{ ( {x}^{2} + 5x) \bigg( \dfrac{1}{x} \bigg) - (lnx) (2 \times {x}^{2 - 1} + 5(1))}{ {( {x}^{2} + 5x)}^{2} }}$

${\longmapsto \rm \dfrac{d}{dx}(y) = \dfrac{ ( {x}^{2} + 5x) \bigg( \dfrac{1}{x} \bigg) - (lnx) (2 {x} + 5)}{ {( {x}^{2} + 5x)}^{2} }}$

$\longmapsto \rm \dfrac{d}{dx}(y) = \dfrac{ ( {x} + 5) - (lnx) (2 {x} + 5)}{ {( {x}^{2} + 5x)}^{2} }$

ㅤ

∴ dy/dx of lnx/x² + 5x is (x + 5) - (lnx)(2x + 5)/(x² + 5x)².

More to know :

$\begin{gathered}\boxed{\begin{array}{cc}\bf{\dag}\:\:\underline{\text{Differentiation \:formulae:}}\\\\\bigstar\:\:\sf{ \dfrac{d}{dx} \times { {x}^{n}} = n \times {x}^{n - 1} } \\\\\bigstar\:\:\sf{ \dfrac{d}{dx} ( {x}) = 1 }\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( { Cons}) = 0\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( { log(x) }) = \dfrac{1}{x} \\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\sqrt{x}} )= \dfrac{1}{2 \sqrt{x} }\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\sin(x)} )= \cos(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\cos(x)} )= - \sin(x) \\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\sin(x)} )= \cos(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\tan(x)} )= { \sec}^{2} (x) \\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\cot(x)} )=- { \cosec}^{2} (x) \\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\sec(x)} )=\sec(x) \times \tan(x)\\\\\bigstar\:\:\sf\dfrac{d}{dx} ( {\cosec(x)})= - \cosec(x) \times \cot(x)\end{array}}\end{gathered}$