Question

If f(x) = 3√x in(x), find f'(x).

b)Find f'(4)​

Answer 1

Answer:

6

Step-by-step explanation:

f(x)= 3rootx

Now putting x = 4

f(4)= 3root4

f(4)= 3×2

f(4)= 6

Please mark me as brainliest

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:f(x) = 3 \sqrt{x} \: ln(x)$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) =\dfrac{d}{dx}( 3 \sqrt{x} \: ln(x))$

$\rm :\longmapsto\:f'(x) = 3\bigg(\dfrac{d}{dx} \sqrt{x} \: ln(x) \bigg)$

$\rm :\longmapsto\:f'(x) = 3\bigg(ln(x) \: \dfrac{d}{dx} \sqrt{x} \: + \: \sqrt{x}\dfrac{d}{dx} \: ln(x) \bigg)$

$\red{\bigg \{ \because \: \dfrac{d}{dx}u.v \: = \: v \: \dfrac{d}{dx}u \: + \: u \: \dfrac{d}{dx}v\bigg \}}$

$\rm :\longmapsto\:f'(x) = 3\bigg(ln(x) \: \dfrac{1}{2 \sqrt{x} } + \sqrt{x} \times \dfrac{1}{x} \bigg)$

$\red{\bigg \{ \sf\because \: \dfrac{d}{dx}ln(x) = \dfrac{1}{x} \bigg \}} \\ \red{\bigg \{ \sf \because \:\dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} } \bigg \}}$

$\rm :\longmapsto\:f'(x) = 3\bigg(\: \dfrac{ln(x)}{2 \sqrt{x} } + \dfrac{1}{ \sqrt{x} } \bigg)$

$\rm :\longmapsto\:f'(4) = 3\bigg(\: \dfrac{ln(4)}{2 \sqrt{4} } + \dfrac{1}{ \sqrt{4} } \bigg)$

$\rm :\longmapsto\:f'(4) = 3\bigg(\: \dfrac{2 \: ln(2)}{2 \times 2 } + \dfrac{1}{2 } \bigg)$

$\rm :\longmapsto\:f'(4) = 3\bigg(\: \dfrac{\: ln(2)}{2} + \dfrac{1}{2 } \bigg)$

$\rm :\longmapsto\:f'(4) = 3\bigg(\: \dfrac{\: ln(2) + 1}{2}\bigg)$

Additional Information :-

$\rm :\longmapsto\:\dfrac{d}{dx}k = 0$

$\rm :\longmapsto\:\dfrac{d}{dx}x = 1$

$\rm :\longmapsto\:\dfrac{d}{dx}sinx = cosx$

$\rm :\longmapsto\:\dfrac{d}{dx}cosx = - \: sinx$

$\rm :\longmapsto\:\dfrac{d}{dx}cosecx = - \: cosecx \: cotx$

$\rm :\longmapsto\:\dfrac{d}{dx}secx = \: secx \: tanx$

$\rm :\longmapsto\:\dfrac{d}{dx}tanx = \: {sec}^{2}x$

$\rm :\longmapsto\:\dfrac{d}{dx}cotx = - \: {cosec}^{2}x$