What is the derivative of #f(x)=x*log_5(x)# ?

Question

Ezra Adams · Accepted Answer

When you're differentiating an exponential with a base other than #e#, use the change-of-base rule to convert it to natural logarithms:

#f(x) = x * lnx/ln5#

Now, differentiate, and apply the product rule:

#d/dxf(x) = d/dx[x] * lnx/ln5 + x * d/dx[lnx/ln5]#

We know that the derivative of #ln x# is #1/x#. If we treat #1/ln5# as a constant, then we can reduce the above equation to:

#d/dxf(x) = lnx/ln5 + x/(xln5)#

Simplifying yields:

#d/dxf(x) = (lnx+1)/ln5#

Scarlett Allison · Answer

undefined To find the derivative of $ f(x) = x \cdot \log_5(x) $, you can use the product rule. The product rule states that if you have two functions, $ u(x) $ and $ v(x) $, then the derivative of their product is given by $ (u \cdot v)' = u'v + uv' $.

Let $ u(x) = x $ and $ v(x) = \log_5(x) $.

The derivative of $ u(x) = x $ is $ u'(x) = 1 $.

The derivative of $ v(x) = \log_5(x) $ can be found using the chain rule:

$$ v'(x) = \frac{1}{x \ln(5)} $$

Now, applying the product rule:

$$ f'(x) = u'v + uv' $$
$$ f'(x) = 1 \cdot \log_5(x) + x \cdot \frac{1}{x \ln(5)} $$
$$ f'(x) = \log_5(x) + \frac{1}{\ln(5)} $$

So, the derivative of $ f(x) = x \cdot \log_5(x) $ is $ \log_5(x) + \frac{1}{\ln(5)} $.