What is the derivative of #y=log_10x/x#?

Question

Ezekiel Alexander · Accepted Answer

#(dy)/(dx)=log_10e((1-lnx)/x^2)=0.4343((1-lnx))/x^2# #y=log_10x/x# can be written as #y=log_10 exxlnx/x# or #y=0.4343xxlnx/x# and using quotient rule Hence #(dy)/(dx)=0.4343xx(x xx1/x-lnx xx1)/x^2# = #0.4343((1-lnx))/x^2#

Isaiah Allen · Answer

undefined To find the derivative of $ y = \frac{\log_{10} x}{x} $, you can use the quotient rule.

Let $ u = \log_{10} x $ and $ v = x $.

Then, $ u' = \frac{1}{x \ln 10} $ and $ v' = 1 $.

Using the quotient rule:

$$ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} $$

$$ = \frac{\frac{1}{x \ln 10} \cdot x - \log_{10} x \cdot 1}{x^2} $$

$$ = \frac{1}{x \ln 10} - \frac{\log_{10} x}{x^2} $$

Therefore, the derivative of $ y = \frac{\log_{10} x}{x} $ is $ \frac{1}{x \ln 10} - \frac{\log_{10} x}{x^2} $.

Isaiah Allen · Answer

undefined To find the derivative of $ y = \frac{\log_{10}x}{x} $, you can use the quotient rule:

Given $ y = \frac{\log_{10}x}{x} $,

The derivative is given by:

$$ \frac{d}{dx}\left(\frac{\log_{10}x}{x}\right) = \frac{\frac{d}{dx}(\log_{10}x) \cdot x - \log_{10}x \cdot \frac{d}{dx}(x)}{x^2} $$

Using the chain rule for differentiation, $\frac{d}{dx}(\log_{10}x) = \frac{1}{x \ln(10)} $.

And $\frac{d}{dx}(x) = 1$.

So, plugging into the formula, we get:

$$ \frac{1}{x \ln(10)} \cdot x - \frac{\log_{10}x \cdot 1}{x^2} = \frac{1}{\ln(10)} - \frac{\log_{10}x}{x^2} $$

Thus, the derivative of $ y = \frac{\log_{10}x}{x} $ is $ \frac{1}{\ln(10)} - \frac{\log_{10}x}{x^2} $.