Question

Suppose that f(x) = 18e^x+7ln(x). Find f'(3).​

Answer 1

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

Given that

$\rm :\longmapsto\:f(x) = 18 {e}^{x} + 7ln \: x$

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

$\rm :\longmapsto\:\dfrac{d}{dx} f(x) =\dfrac{d}{dx}( 18 {e}^{x} + 7ln \: x)$

$\rm :\longmapsto\:f'(x) =\dfrac{d}{dx}18 {e}^{x} + \dfrac{d}{dx}7 \: ln \: x$

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

$\rm :\longmapsto\:f'(x) =18\dfrac{d}{dx} {e}^{x} +7 \dfrac{d}{dx}\: ln \: x$

$\: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: \dfrac{d}{dx}k \: f(x) = k \: \dfrac{d}{dx}f(x)\bigg \}}$

$\rm :\longmapsto\:f'(x) = 18 {e}^{x} \: + \: 7 \times \dfrac{1}{x}$

$\red{\bigg \{ \because \:\dfrac{d}{dx} {e}^{x} = {e}^{x} \: \: \: \: and \: \: \: \: \dfrac{d}{dx}lnx = \dfrac{1}{x} \bigg \}}$

$\rm :\longmapsto\:f'(x) = 18 {e}^{x} \: + \: \dfrac{7}{x}$

Thus,

$\rm :\longmapsto\:f'(3) = 18 {e}^{3} \: + \: \dfrac{7}{3}$

Additional Information :-

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

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

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

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

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

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

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

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

Answer 2

Answer:

f'(3) = 18e^3 + 7/3 = 363.87

Step-by-step explanation:

f(x) = 18*exp(x) + 7ln(x)

f'(x) = 18*{exp(x)}' + 7{ln(x)}'

f'(x) = 18*exp(x) + 7/x

f'(3) = 18*exp(3) + 7/3

f'(3) = 18*e^3 + 7/3 = 363.87

Prerequisites :

• (a+b)' = a' + b'
• {exp(x)}' = exp(x)
• {ln(x)}' = 1/x