Calculus with Different Bases

Expert reviewed • 22 November 2024 • 6 minute read

HSC Maths Advanced Syllabus

establish and use the formula $\frac{d}{dx}(log_ax)=\frac{1}{xlna}$
establish and use the formula $\frac{d}{dx}(a^x)=(lna)a^x$ $\frac{d}{d x} (a^{x}) = (l na) a^{x}$
- using graphing software or otherwise, sketch and explore the gradient function for a given exponential function, recognise it as another exponential function and hence determine the relationship between exponential functions and their derivatives

Note:

Video coming soon!

Logarithmic Functions With Other Bases

So far in this module, we have not discussed how to use calculus, when a logarithmic function has a base other than $e$ . Before we can use any type of formula to derive a function in the form $log_ax$ , we must first learn how to change it’s base. The formula for this is as follows:

log_ax=\frac{log_ex}{log_ea}

Although the change of base formula is useful in deriving a function, a simpler method is to utilise the following formula.

\frac{d}{dx}log_ax=\frac{1}{xlog_ea}

Practice Question 1

Differentiate the following function: $y=log_2x+4log_62x$

Applying the formula above, we can determine the derivative of this function:

y'=\frac{1}{1\times log_e2}+4\times \frac{1}{x\times log_e6}\\=\frac{1}{xlog_e2}+\frac{4}{xlog_e6}\\

Exponential Functions With Other Bases

Additionally in this module, we have not explored using calculus with exponential functions that have bases other than $e$ . Before we can use any formula to derive a function in the form $a^x$ , we must first revise how to change its base.

a^x=e^{xlog_ea}

Now that we know how to change the base of a function in the form $a^x$ , we can apply calculus to these functions in two different ways.

Convert all expressions to be in a form where $e$ is the base, and use previously learned formulas to differentiate or integrate.
Use the following formulas:

\frac{d}{dx}a^x=a^xlog_ea\quad and \quad \int a^xdx=\frac{a^x}{log_ea}+C

Practice Question 2

Differentiate the following function: $f(x)=8-2^x$

Using the formulas provided above, we can derive this function by substituting terms into the formula.

f'(x)=-(2^x\times log_e2)\\=-2^xlog_e2

In this case the negative sign is not attached to the value of 2, and thus is treated as a multiplication factor of the derivative.

Practice Question 3

Solve the following expression: $\int_0^2 5^xdx$

Using the formulas provided above and previously learned knowledge, we can evaluate the given definite integral.

\int_0^25^xdx=[\frac{5^x}{log_e5}]_0^2\\=(\frac{5^2}{log_e5})-(\frac{5^0}{log_e5})\\=(\frac{25}{log_e5})-(\frac{1}{log_e5})\\=\frac{24}{log_e5}

Integration of the Reciprocal Function

Return to Exponential and Logarithmic Functions