How do you differentiate #y=ln(x^2+1)-log_2(5x)#?

Question

Charles Anctil · Accepted Answer

Please see below.

#d/dx(ln(x^2+1))# use #d/dx(lnu) = 1/u (du)/dx# so #d/dx(ln(x^2+1)) = (2x)/(x^2+1)# #d/dx(log_2(5x)) = d/dx(log_2(5))+d/dx(log_2(x))# # = 0 + d/dx(log_2(x))# # = 1/(xln2)# So, for #y = ln(x^2+1)-log_2(5x)#, we have #dy/dx = (2x)/(x^2+1)-1/(xln2)# Note #log_2(x) = lnx/ln2 = 1/ln2 * lnx# So the derivtaive is #1/ln2 * 1/x = 1/(xln2)#

Christopher Allers · Answer

undefined To differentiate the function $ y = \ln(x^2 + 1) - \log_2(5x) $, you would use the rules of differentiation. The derivative of $ \ln(u) $ with respect to $ x $ is $ \frac{1}{u} \frac{du}{dx} $, and the derivative of $ \log_a(v) $ with respect to $ x $ is $ \frac{1}{v} \frac{dv}{dx} \log_a(e) $. Applying these rules:

$$
\frac{dy}{dx} = \frac{1}{x^2 + 1} \cdot \frac{d(x^2 + 1)}{dx} - \frac{1}{5x} \cdot \frac{d(5x)}{dx} \log_2(e)
$$

Now, differentiate $ x^2 + 1 $ and $ 5x $:

$$
\frac{d(x^2 + 1)}{dx} = 2x \quad \text{and} \quad \frac{d(5x)}{dx} = 5
$$

So, substituting these derivatives back into the equation:

$$
\frac{dy}{dx} = \frac{1}{x^2 + 1} \cdot 2x - \frac{1}{5x} \cdot 5 \log_2(e)
$$

Simplify:

$$
\frac{dy}{dx} = \frac{2x}{x^2 + 1} - \frac{\log_2(e)}{x}
$$

This is the derivative of the function $ y = \ln(x^2 + 1) - \log_2(5x) $.