Question

13. if f(x) = x2 + loge x + loge 5 then f' (1) is
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Answer 1

Answer:

3

Step-by-step explanation:

Given,

$\sf \bf :\longmapsto f(x) = x^2 + log_ex + log_e5$

Bow finding the derivative, use the formula:

$\boxed{\boxed{\begin{array}{l}\dfrac{d}{dx} log_ex = \dfrac{1}{x}\\ \\ \hline \\ \dfrac{d}{dx}x^n = nx^{n-1}\end{array}}}$

$\sf \bf :\longmapsto f'(x) = 2x + \dfrac{1}{x} + 0$

$\sf \bf :\longmapsto f'(x) = 2x + \dfrac{1}{x}$

$\sf \bf :\longmapsto f'(1) = 2(1) + \dfrac{1}{1}$

$\sf \bf :\longmapsto f'(1) = 2 + 1 = 3$

Additional Information:

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Answer:

here's the right ans hope it's useful as follows:::