Question

find the differentiation of 2x^5+49x+3x+1/5√x+3/x
with full explanation​

Answer 1

GIVEN :–

• An equation –

$\\ \implies\bf 2x^5+ 49x + 3x + \dfrac{1}{5 \sqrt{x}} +\dfrac{3}{x}$

TO FIND :–

• Differentiation = ?

SOLUTION :–

• Let the function –

$\\ \implies\bf P = 2x^5+ 49x + 3x + \dfrac{1}{5 \sqrt{x}} +\dfrac{3}{x}$

$\\ \implies\bf P = 2x^5+ (49 + 3)x + \dfrac{1}{5 {x}^{ \frac{1}{2} } } +\dfrac{3}{x}$

$\\ \implies\bf P = 2x^5+52x + \dfrac{{x}^{ - \frac{1}{2} } }{5} +3 {x}^{ - 1}$

• We know that –

$\\ \longrightarrow\bf \pink{\dfrac{d{x}^{n}}{dx} = n {x}^{n - 1}}$

• So that –

$\\ \implies\bf \dfrac{dP}{dx} = \dfrac{d}{dx} \bigg\{ 2x^5+52x + \dfrac{{x}^{ - \frac{1}{2} } }{5} +3 {x}^{ - 1} \bigg\}$

$\\ \implies\bf \dfrac{dP}{dx} = \dfrac{d}{dx}(2x^5)+\dfrac{d}{dx}(52x) +\dfrac{d}{dx} \bigg(\dfrac{{x}^{ - \frac{1}{2} } }{5} \bigg)+\dfrac{d}{dx}(3 {x}^{ - 1})$

$\\ \implies\bf \dfrac{dP}{dx} =2 \dfrac{d}{dx}(x^5)+52\dfrac{d}{dx}(x) + \dfrac{1}{5} \dfrac{d}{dx}({x}^{ - \frac{1}{2} })+3\dfrac{d}{dx}( {x}^{ - 1})$

$\\ \implies\bf \dfrac{dP}{dx} =2(5 {x}^{4})+52+ \dfrac{1}{5} \times - \dfrac{1}{2} ({x}^{ - \frac{1}{2} - 1})+3( - 1)( {x}^{ - 1 - 1})$

$\\ \implies\bf \dfrac{dP}{dx} =10 {x}^{4}+52 - \dfrac{1}{10}({x}^{ - \frac{3}{2} }) - 3( {x}^{ -2})$

$\\ \implies \large \pink{ \boxed{\bf \dfrac{dP}{dx} =10 {x}^{4}+52 - \dfrac{1}{10x \sqrt{x}}- \dfrac{3}{ {x}^{2}}}}$

Answer 2

Step-by-step explanation:

Given Equation:-

$\sf 2x^5+49x+3x+\dfrac {1}{5\sqrt {3}x+\dfrac{3}{x}$

To do:-

Find differentiation

Solution:-

Let the function be P

$\\ \qquad {:}\longrightarrow\sf P = 2x^5+ 49x + 3x + \dfrac{1}{5 \sqrt{x}} +\dfrac{3}{x} \\ \\\qquad \implies\sf P = 2x^5+ (49 + 3)x + \dfrac{1}{5 {x}^{ \frac{1}{2} } } +\dfrac{3}{x} \\ </p><p></p><p>\\ \qquad {:}\implies\sf P = 2x^5+52x + \dfrac{{x}^{ - \frac{1}{2} } }{5} +3 {x}^{ - 1}$

As we know that

$\boxed { \rightarrowtail\sf {\dfrac{d{x}^{n}}{dx} = n {x}^{n - 1}} }$

$\\\qquad {:} \implies\sf \dfrac{dP}{dx} = \dfrac{d}{dx} \bigg\{ 2x^5+52x + \dfrac{{x}^{ - \frac{1}{2} } }{5} +3 {x}^{ - 1} \bigg\}$

$\\\qquad {:} \implies\sf \dfrac{dP}{dx} = \dfrac{d}{dx}(2x^5)+\dfrac{d}{dx}(52x) +\dfrac{d}{dx} \bigg(\dfrac{{x}^{ - \frac{1}{2} } }{5} \bigg)+\dfrac{d}{dx}(3 {x}^{ - 1})$

$\\\qquad {:} \implies\sf \dfrac{dP}{dx} =2 \dfrac{d}{dx}(x^5)+52\dfrac{d}{dx}(x) + \dfrac{1}{5} \dfrac{d}{dx}({x}^{ - \frac{1}{2} })+3\dfrac{d}{dx}( {x}^{ - 1})$

$\\\qquad {:} \implies\sf \dfrac{dP}{dx} =2(5 {x}^{4})+52+ \dfrac{1}{5} \times - \dfrac{1}{2} ({x}^{ - \frac{1}{2} - 1})+3( - 1)( {x}^{ - 1 - 1})$

$\\\qquad {:} \implies\sf \dfrac{dP}{dx} =10 {x}^{4}+52 - \dfrac{1}{10}({x}^{ - \frac{3}{2} }) - 3( {x}^{ -2})$

$\\ \implies\red {\bigstar} \LARGE{ \boxed{\sf \dfrac{dP}{dx} =10 {x}^{4}+52 - \dfrac{1}{10x \sqrt{x}}- \dfrac{3}{ {x}^{2}}}}\red {\bigstar}$