Question

In each of the following

$find \: \frac{dy}{dx}\:$
$a)\: \: y = \sin ^ - 1 \sqrt{x} + ln( \sqrt{x} )$

$b) {e}^{y} = {2}^{x} log(x) - {e}^{2x}$
​

Answer 1

(a) Recall that

$\dfrac{\mathrm d[\sin^{-1}x]}{\mathrm dx}=\dfrac1{\sqrt{1-x^2}}$

$\dfrac{\mathrm d[\ln x]}{\mathrm dx}=\dfrac1x$

$\dfrac{\mathrm d[\sqrt x]}{\mathrm dx}=\dfrac1{2\sqrt x}$

Then by the chain rule,

$y=\sin^{-1}(\sqrt x)+\ln(\sqrt x)$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d[\sin^{-1}(\sqrt x)+\ln(\sqrt x)]}{\mathrm dx}$

$=\dfrac1{\sqrt{1-(\sqrt x)^2}}\dfrac{\mathrm d[\sqrt x]}{\mathrm dx}+\dfrac1{\sqrt x}\dfrac{\mathrm d[\sqrt x]}{\mathrm dx}$

$=\dfrac1{2\sqrt x\sqrt{1-x}}+\dfrac1{2(\sqrt x)^2}$

$=\dfrac1{2\sqrt{x-x^2}}+\dfrac1{2x}$

$=\boxed{\dfrac{x+\sqrt{x-x^2}}{2x\sqrt{x-x^2}}}$

(b) I'll assume $\log x$ refers to the logarithm of base $e$. Recall that

$\dfrac{\mathrm d[e^x]}{\mathrm dx}=e^x$

$2^x=e^{\log2\,x}\implies\dfrac{\mathrm d[2^x]}{\mathrm dx}=e^{\log2\,x}\dfrac{\mathrm d[\log2\,x]}{\mathrm dx}=\log2\cdot2^x$

Then using the chain rule and implicit differentiation, we get

$e^y=2^x\log x-e^{2x}$

$\implies e^y\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d[2^x]}{\mathrm dx}\log x+2^x\dfrac{\mathrm d[\log x]}{\mathrm dx}-e^{2x}\dfrac{\mathrm d[2x]}{\mathrm dx}$

$e^y\dfrac{\mathrm dy}{\mathrm dx}=\log2\cdot2^x\log x+\dfrac{2^x}x-2e^{2x}$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\log2\cdot2^x\log x+\frac{2^x}x-2e^{2x}}{e^y}$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\log2\cdot2^x\log x+\frac{2^x}x-2e^{2x}}{2^x\log x-e^{2x}}$

$\dfrac{\mathrm dy}{\mathrm dx}=\boxed{\dfrac{\log2\cdot2^xx\log x+2^x-2xe^{2x}}{x(2^x\log x-e^{2x})}}$

If instead $\log x=\log_{10}x$ denotes the base 10 logarithm, you need only make the following adjustment:

$\log x=\dfrac{\ln x}{\ln 10}\implies\dfrac{\mathrm d[\log_{10}x]}{\mathrm dx}=\dfrac1{\ln10\,x}$

where $\ln x$ is logarithm of base $e$.