Question

f(x)= (10^x +logx)/√x find f'(x)=?

Answer 1

$$f(x)={10^x+\log(x) \over \sqrt{x}}={10^x+\ln(x) \over \sqrt{x}}$$

Find $f'(x)$

$$f'(x)=\big({10^x+\ln(x) \over \sqrt{x}}\big)'=$$

$$={(10^x+\ln(x))'\sqrt{x} -(\sqrt{x})'(10^x+\ln(x))\over( \sqrt{x})^2}=$$

$$={\ln(10)\cdot10^x\sqrt{x}+{1 \over x}\cdot\sqrt{x} -{1 \over 2\sqrt{x}}(10^x+\ln(x))\over x}=$$

$$={2\ln(10)\cdot x\cdot10^x+2 -10^x-\ln(x)\over 2x^{{3 \over 2}}}$$

$$f'(x)={2\ln(10)\cdot x\cdot10^x+2 -10^x-\ln(x)\over 2x^{{3 \over 2}}}$$