Finding the Derivative of \({{\log }_{10}}\left( 5{{\rm{x}}^2} + 3 \right)\)

This question asks us to find the derivative of a logarithmic function with a base other than \(e\). The function is given as \(y = {{\log }_{10}}\left( 5{{\rm{x}}^2} + 3 \right)\) with respect to \(x\).

Understanding Logarithmic Derivatives

The standard derivative rule for a logarithm with base \(b\) is:

\(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\log }_b}\left( u \right)} \right) = \frac{1}{{u \cdot \ln \left( b \right)}} \cdot \frac{{\rm{du}}}{{{\rm{dx}}}}\)

Where \(u\) is a function of \(x\), and \(\ln(b)\) is the natural logarithm of the base \(b\).

Applying the Derivative Rule and Chain Rule

In our function \(y = {{\log }_{10}}\left( 5{{\rm{x}}^2} + 3 \right)\):

• The base \(b\) is 10.
• The argument \(u\) is \(5{{\rm{x}}^2} + 3\).

We need to find the derivative of \(u\) with respect to \(x\):

\(\frac{{\rm{du}}}{{{\rm{dx}}}} = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( 5{{\rm{x}}^2} + 3 \right)\)

Using the power rule \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( c{{x}^n} \right) = cn{{x}}^{n-1}\) and the rule for the derivative of a constant \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( c \right) = 0\):

\(\frac{{\rm{du}}}{{{\rm{dx}}}} = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( 5{{\rm{x}}^2} \right) + \frac{{\rm{d}}}{{{\rm{dx}}}}\left( 3 \right) = 5 \cdot 2{{\rm{x}}^{2-1}} + 0 = 10{\rm{x}}\)

Now, we substitute \(u = 5{{\rm{x}}^2} + 3\), \(\frac{{\rm{du}}}{{{\rm{dx}}}} = 10{\rm{x}}\), and \(b = 10\) into the general logarithmic derivative formula:

\(\frac{{\rm{dy}}}{{{\rm{dx}}}} = \frac{1}{{\left( 5{{\rm{x}}^2} + 3 \right) \cdot \ln \left( 10 \right)}} \cdot 10{\rm{x}}\)

\(\frac{{\rm{dy}}}{{{\rm{dx}}}} = \frac{{10{\rm{x}}}}{{\left( 5{{\rm{x}}^2} + 3 \right) \cdot \ln \left( 10 \right)}}\)

Expressing in terms of \({{\log }_{10}}{\rm{e}}\)

The given options involve \({{\log }_{10}}{\rm{e}}\). We need to relate \(\frac{1}{{\ln \left( 10 \right)}}\) to \({{\log }_{10}}{\rm{e}}\).

Recall the change of base formula for logarithms: \({{\log }_b}a = \frac{{{{\log }_c}a}}{{{{\log }_c}b}}\). Using the natural logarithm (\(c=e\)):

\({{\log }_{10}}{\rm{e}} = \frac{{{{\log }_e}{\rm{e}}}}{{{{\log }_e}10}}\)

Since \({{\log }_e}{\rm{e}} = \ln \left( {\rm{e}} \right) = 1\) and \({{\log }_e}10 = \ln \left( 10 \right)\), we have:

\({{\log }_{10}}{\rm{e}} = \frac{1}{{\ln \left( 10 \right)}}\)

Now, substitute \(\frac{1}{{\ln \left( 10 \right)}}\) with \({{\log }_{10}}{\rm{e}}\) in our derivative expression:

\(\frac{{\rm{dy}}}{{{\rm{dx}}}} = \frac{{10{\rm{x}}}}{{\left( 5{{\rm{x}}^2} + 3 \right)}} \cdot \frac{1}{{\ln \left( 10 \right)}} = \frac{{10{\rm{x}}}}{{\left( 5{{\rm{x}}^2} + 3 \right)}} \cdot {{\log }_{10}}{\rm{e}}\)

So, the derivative is:

\(\frac{{\rm{dy}}}{{{\rm{dx}}}} = \frac{{10{\rm{x}}{{\log }_{10}}{\rm{e}}}}{{{5{\rm{x}}^2} + 3}}\)

Step-by-Step Solution

1. Identify the function: \(y = {{\log }_{10}}\left( 5{{\rm{x}}^2} + 3 \right)\).
2. Identify the argument \(u = 5{{\rm{x}}^2} + 3\) and the base \(b=10\).
3. Calculate the derivative of \(u\) with respect to \(x\): \(\frac{{\rm{du}}}{{{\rm{dx}}}} = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( 5{{\rm{x}}^2} + 3 \right) = 10{\rm{x}}\).
4. Apply the logarithm derivative rule \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\log }_b}\left( u \right)} \right) = \frac{1}{{u \cdot \ln \left( b \right)}} \cdot \frac{{\rm{du}}}{{{\rm{dx}}}}\).
5. Substitute the values: \(\frac{{\rm{dy}}}{{{\rm{dx}}}} = \frac{1}{{\left( 5{{\rm{x}}^2} + 3 \right) \cdot \ln \left( 10 \right)}} \cdot 10{\rm{x}}\).
6. Simplify and rewrite using \({{\log }_{10}}{\rm{e}} = \frac{1}{{\ln \left( 10 \right)}}\).
7. The final derivative is \(\frac{{10{\rm{x}}{{\log }_{10}}{\rm{e}}}}{{{5{\rm{x}}^2} + 3}}\).

Comparison with Options

Let's compare our derived derivative with the given options:

The calculated derivative matches Option 3.

Revision Table: Calculus Derivatives

Additional Information: Logarithm Properties and Calculus

Understanding logarithm properties is crucial for calculus involving logarithmic functions. The change of base formula is particularly useful when dealing with logarithms not in base \(e\) or base 10, or when converting between different bases like we did here with \(\ln(10)\) and \({{\log }_{10}}{\rm{e}}\).

• Change of Base Formula: \({{\log }_b}a = \frac{{{{\log }_c}a}}{{{{\log }_c}b}}\). Common choices for \(c\) are \(e\) (natural log, \(\ln\)) or 10 (common log, \({{\log }_{10}}\)).
• Relationship between \(\ln(b)\) and \({{\log }_b}{\rm{e}}\): As shown above, \({{\log }_b}{\rm{e}} = \frac{1}{{\ln (b)}}\). This identity helps in simplifying expressions involving different bases.
• Chain Rule: When differentiating a composite function like \({{\log }_{10}}(g(x))\), where \(g(x) = 5{{\rm{x}}^2} + 3\), the chain rule states that the derivative is the derivative of the outer function (log base 10) evaluated at the inner function, multiplied by the derivative of the inner function. This is exactly what the formula \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\log }_b}\left( u \right)} \right) = \frac{1}{{u \cdot \ln \left( b \right)}} \cdot \frac{{\rm{du}}}{{{\rm{dx}}}}\) represents.