Economics at Zero Marginal Cost: Differentiation by taking Logarithmic

Exercise:

Find the derivatives of:

$\\1.\; y=x^x\\ 2.\;y=b^x\\ 3.\;y=x^{1/x}\\ 4.\;y=x^{x+4}\\ 5.\;y=(x+1)^x\\ 6.\;y=(X-1)^{x+1}\\ 7.\;y=log_{b}\;x\\ 8.\;f(x)=x^6+log_{a}\;x$

Exercise:

1. $y=x^x$

Solution:

$y=x^x$

Taking $log$ on both sides,

$log\;y=log\;x^x$

$log\;y=x\;log\;x$

Differentiating both sides with respect to $x$

$\frac{d}{dx}y=\frac{d}{dx}x\;log\;x$

$\\\frac{1}{y}.\frac{d}{dx}y=x\frac{d}{dx}log\;x+log\;x\frac{d}{dx}x\\ \frac{1}{y}.\frac{dy}{dx}=x\frac{1}{x}+log\;x.1\\ \frac{1}{y}.\frac{dy}{dx}=1+log\;x\\ \frac{dy}{dx}=\frac{1+log\;x}{\frac{1}{y}}\\ \frac{dy}{dx}=y(1+log\;x)\\ \frac{dy}{dx}=x^x(1=log\;x)$

2. $y=b^x$

Solution:

$y=b^x$

Taking $log$ on both sides,

$log\;y=log\;b^x$

$log\;y=x\;log\;b$

Differentiating both sides with respect to $x$

$\frac{d}{dx}log\;y=\frac{d}{dx}x\;log\;b$

$\\\frac{1}{y}.\frac{d}{dx}y=log\;b\frac{d}{dx}x\\ \frac{1}{y}.\frac{dy}{dx}=log\;b.1\\ \frac{1}{y}.\frac{dy}{dx}=log\;b\\ \frac{dy}{dx}=\frac{log\;b}{\frac{1}{y}}\\ \frac{dy}{dx}=y\;log\;b\\ \frac{dy}{dx}=b^xlog\;b$

3. $y=x^{1/x}$

Solution:

$y=x^{1/x}$

Taking $log$ on both sides,

$\\log\;y=log\;x^{1/x}\\ log\;y=\frac{1}{x}log\;x$

Differentiating both sides with respect to $x$

$\frac{d}{dx}log\;y=\frac{d}{dx}\frac{1}{x}logx$

$\\\frac{1}{y}\frac{d}{dx}y=\frac{1}{x}\frac{d}{dx}log\;x+log\;x\frac{d}{dx}\frac{1}{x}\\ \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}.\frac{1}{x}+log\;x.\frac{d}{dx}x^{-1}\\ \frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2}+log\;x.-1x^{-1-1}\\ \frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2}+log\;x.x^{-2}\\ \frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2}-log\;x.\frac{1}{x^2}\\ \frac{dy}{dx}=\frac{\frac{1}{x^2}-log\;x.\frac{1}{x^2}}{\frac{1}{y}}\\ \frac{dy}{dx}=y\left ( \frac{1}{x^2}-log\;x.\frac{1}{x^2} \right )\\ \frac{dy}{dx}=x^{1/x}\left ( \frac{1}{x^2}-logx.\frac{1}{x^2} \right )$

4. $y=x^{x+4}$

Solution:

$y=x^{x+4}$

Taking $log$ on both sides,

$\\log\;y=log\;x^{x+4}\\ log\;y=(x+4)log\;x$

Differentiating both sides with respect to $x$

$\frac{d}{dx}log\;y=\frac{d}{dx}(x+4)log\;x$

$\\\frac{1}{y}\frac{d}{dx}y=(x+4)\frac{d}{dx}log\;x+log\;x\frac{d}{dx}(x+4)\\ \frac{1}{y}\frac{dy}{dx}=(x+4).\frac{1}{x}+log\;x\left ( \frac{d}{dx}x+\frac{d}{dx}4 \right )\\ \frac{1}{y}\frac{dy}{dx}=\frac{x+4}{x}+log\;x(1+0)\\ \frac{1}{y}\frac{dy}{dx}=\frac{x+4}{x}+log\;x.1\\ \frac{1}{y}\frac{dy}{dx}=\frac{x+4}{x}+log\;x\\ \frac{dy}{dx}=\frac{\frac{x+4}{x}+log\;x}{\frac{1}{y}}\\ \frac{dy}{dx}=y\left ( \frac{x+4}{x}+log\;x \right )\\ \frac{dy}{dx}=x^{x+4}\left ( \frac{x+4}{x}+log\;x \right )$

5. $y=(x+1)^x$

Solution:

$y=(x+1)^x$

Taking $log$ on both sides,

$\\\log\;y=log\;(x+1)^x\\ log\;y=x\;log\;(x+1)$

Differentiating both sides with respect to $x$

$\frac{d}{dx}log\;y=\frac{d}{dx}x\;log(x+1)$

$\\\frac{1}{y}\frac{d}{dx}y=x\frac{d}{dx}log\;(x+1)+log\;(x+1)\frac{d}{dx}x\\ \frac{1}{y}\frac{dy}{dx}=x.\frac{1}{x+1}\frac{d}{dx}(x+1)+log(x+1).1\\ \frac{1}{y}\frac{dy}{dx}=\frac{x}{x+1}\left ( \frac{d}{dx}x+\frac{d}{dx}1 \right )+log(x+1)\\ \frac{1}{y}\frac{dy}{dx}=\frac{x}{x+1}(1+0)+log(x+1)\\ \frac{1}{y}\frac{dy}{dx}=\frac{x}{x+1}.1+log(x+1)\\ \frac{dy}{dx}=\frac{x}{x+1}+log(x+1)\\ \frac{dy}{dx}=\frac{\frac{x}{x+1}+log(x+1)}{\frac{1}{y}}\\ \frac{dy}{dx}=y(\frac{x}{x+1}+log(x+1))\\ \frac{dy}{dx}=(x+1)^x(\frac{x}{x+1}+log(x+1))$

6. $y=(x-1)^{x+1}$

Solution:

$y=(x-1)^{x+1}$

Taking $log$ on both sides,

$\\\log\;y=log\;(x-1)^{x+1}\\ log\;y=(x+1)\;log\;(x-1)$

Differentiating both sides with respect to $x$

$\frac{d}{dx}log\;y=\frac{d}{dx}(x+1)\;log(x-1)$

$\\\frac{1}{y}\frac{d}{dx}y=(x+1)\frac{d}{dx}log(x-1)+log(x-1)\frac{d}{dx}(x+1)\\ \frac{1}{y}\frac{dy}{dx}=(x+1).\frac{1}{x-1}\frac{d}{dx}(x-1)+log(x-1)\left ( \frac{d}{dx}x+\frac{d}{dx}1 \right )\\ \frac{1}{y}\frac{dy}{dx}=\frac{x+1}{x-1}\left ( \frac{d}{dx}x-\frac{d}{dx}1 \right )+log(x-1)(1+0)\\ \frac{1}{y}\frac{dy}{dx}=\frac{x+1}{x-1}(1-0)+log(x-1).1\\ \frac{1}{y}\frac{dy}{dx}=\frac{x+1}{x-1}.1+log(x-1)\\ \frac{1}{y}\frac{dy}{dx}=\frac{x+1}{x-1}+log(x-1)\\ \frac{dy}{dx}=\frac{\frac{x+1}{x-1}+log(x-1)}{\frac{1}{y}}\\ \frac{dy}{dx}=y\left ( \frac{x+1}{x-1}+log(x-1) \right )\\ \frac{dy}{dx}=(x-1)^{x+1}\left ( \frac{x+1}{x-1}+log(x-1) \right )$

7. $y=log\;b^x$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}(log\;b^x)\\ \frac{dy}{dx}=\frac{d}{dx}\left ( \frac{log\;x}{log\;b} \right )\\ \frac{dy}{dx}=\frac{d}{dx}\left ( \frac{1}{log\;b}.log\;x \right )\\ \frac{dy}{dx}=\frac{1}{log\;b}.\frac{d}{dx}log\;x\\ \frac{dy}{dx}=\frac{1}{log\;b}.\frac{1}{x}\\ \frac{dy}{dx}=\frac{1}{(log\;b)x}\\$

8. $y=x^6+log\;_{a}\;x$

Solution:

$\\\frac{dy}{dx}=\frac{d}{dx}(x^6+log\;_{a}\;x)\\ \frac{dy}{dx}=\frac{d}{dx}x^6+\frac{d}{dx}log\;_{a}\;x\\ \frac{dy}{dx}=6x^{6-1}+\frac{d}{dx}\left ( \frac{log\;x}{log\;a} \right )\\ \frac{dy}{dx}=6x^5+\frac{d}{dx}\left ( \frac{1}{log\;a}log\;x \right )\\ \frac{dy}{dx}=6x^5+\frac{1}{log\;a}\frac{d}{dx}log\;x\\ \frac{dy}{dx}=6x^5+\frac{1}{log\;a}.\frac{1}{x}\\ \frac{dy}{dx}=6x^5+\frac{1}{(log\;a)x}$

Economics at Zero Marginal Cost

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Wednesday, June 10, 2020

Differentiation by taking Logarithmic

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