![\dfrac{\mathrm d[\sin^{-1}x]}{\mathrm dx}=\dfrac1{\sqrt{1-x^2}}](/tpl/images/0723/4569/5c6d8.png)
![\dfrac{\mathrm d[\ln x]}{\mathrm dx}=\dfrac1x](/tpl/images/0723/4569/07ef4.png)
![\dfrac{\mathrm d[\sqrt x]}{\mathrm dx}=\dfrac1{2\sqrt x}](/tpl/images/0723/4569/8a913.png)
![\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d[\sin^{-1}(\sqrt x)+\ln(\sqrt x)]}{\mathrm dx}](/tpl/images/0723/4569/aaa71.png)
![=\dfrac1{\sqrt{1-(\sqrt x)^2}}\dfrac{\mathrm d[\sqrt x]}{\mathrm dx}+\dfrac1{\sqrt x}\dfrac{\mathrm d[\sqrt x]}{\mathrm dx}](/tpl/images/0723/4569/20cf7.png)
refers to the logarithm of base 
. Recall that![\dfrac{\mathrm d[e^x]}{\mathrm dx}=e^x](/tpl/images/0723/4569/88d07.png)
![2^x=e^{\log2\,x}\implies\dfrac{\mathrm d[2^x]}{\mathrm dx}=e^{\log2\,x}\dfrac{\mathrm d[\log2\,x]}{\mathrm dx}=\log2\cdot2^x](/tpl/images/0723/4569/a69d5.png)
![\implies e^y\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d[2^x]}{\mathrm dx}\log x+2^x\dfrac{\mathrm d[\log x]}{\mathrm dx}-e^{2x}\dfrac{\mathrm d[2x]}{\mathrm dx}](/tpl/images/0723/4569/13f07.png)
denotes the base 10 logarithm, you need only make the following adjustment:![\log x=\dfrac{\ln x}{\ln 10}\implies\dfrac{\mathrm d[\log_{10}x]}{\mathrm dx}=\dfrac1{\ln10\,x}](/tpl/images/0723/4569/2a9a4.png)
is logarithm of base 
In each of the following
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