How do we determine the derivative of #y = log_10sqrt(x^2 + 3x +4)#?

Question

Aurora Alvarado · Accepted Answer

#y = logsqrt(x^2 + 3x + 4)#

#y = 1/2log(x^2 + 3x + 4)#

#2y = log(x^2 + 3x + 4)#

#10^(2y) = x^2 + 3x + 4#

#ln(10^(2y)) = ln(x^2 + 3x + 4)#

#(2y)ln10 = ln(x^2 + 3x + 4)#

Use the chain rule to differentiate the right hand side and the product rule to differentiate the left hand side.

#2(dy/dx)ln10 + 2y(0) = 1/(x^2 + 3x + 4) xx 2x + 3#

#2ln10(dy/dx) = (2x+ 3)/((x + 3)(x + 1))#

#dy/dx = (2x + 3)/(ln100(x+ 3)(x + 1))#

Hopefully this helps!

Aurora Alvarado · Answer

undefined To find the derivative of $ y = \log_{10}\sqrt{x^2 + 3x + 4} $, you can use the chain rule and properties of logarithmic functions. First, rewrite the expression as $ y = \frac{1}{2} \log_{10}(x^2 + 3x + 4) $. Then, differentiate it with respect to $ x $ using the chain rule. The derivative is:

$$ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\ln(10)} \cdot \frac{1}{\sqrt{x^2 + 3x + 4}} \cdot \frac{d}{dx}(x^2 + 3x + 4) $$

$$ \frac{dy}{dx} = \frac{1}{2\ln(10) \sqrt{x^2 + 3x + 4}} \cdot (2x + 3) $$