Differentiation of Exponential and Logarithmic Functions

Expert reviewed 21 July 2024 8 minute read


HSC Maths Advanced Syllabus

  • apply the product, quotient and chain rules to differentiate functions of the form f(x)g(x),f(x)g(x)f(x)g(x),\frac{f(x)}{g(x)} and f(g(x))f(g(x)) where f(x)f(x) and g(x)g(x) are any of the functions covered in the scope of this syllabus, for example xexxe^x, tanxtanx, 1xn\frac{1}{x^n}, xsinxxsinx, exsinxe^{-x}sinx and f(ax+b)f(ax+b)
    • use the composite function rule (chain rule) to establish that ddx[ef(x)]=f(x)eef(x)\frac{d}{dx}[e^{f(x)}]=f'(x)e^{e^{f(x)}}
    • use the composite function rule (chain rule) to establish that ddx[lnf(x)]=f(x)f(x)\frac{d}{dx}[lnf(x)]=\frac{f'(x)}{f(x)}
    • use the logarithmic laws to simplify an expression before differentiating

Note:

Video coming soon!

How to Differentiate Exponential Functions

Building on our knowledge of differentiation from the year 11 course, we should know that there are three standard derivative formulas for exponential functions:

ddxex=ex\frac{d}{dx}e^x=e^x ddxeax+b=aeax+b\frac{d}{dx}e^{ax+b}=ae^{ax+b} ddxeu=eududxorddxef(x)=ef(x)f(x)\frac{d}{dx}e^u=e^u\frac{du}{dx}\quad or \quad \frac{d}{dx}e^{f(x)}=e^{f(x)}f'(x)

It should be noted that the base formula for deriving exe^x is also exe^x. However, when deriving the expression exe^{-x}, it is important to note that the negative sign is moved out the front, as follows: ex-e^{-x}.

Practice Question 1

Derive the following function f(x)=x3e3x+1f(x)=x^3e^{3x+1} using the product rule, and any related formulas provided above.

First we must determine the values of uu and vv.

u=x3v=e3x+1u=x^3\qquad v=e^{3x+1}

Using differentiation formulas and techniques learnt so far, find uu' and vv':

u=3x2v=3e3x+1u'=3x^2\qquad v'=3e^{3x+1}

Finally, we can substitute these values into the product rule formula, to determine the derivative:

f(x)=uv+uv=(3x2×e3x+1)+(x3×3e3x+1)=3x2e3x+1+3x3e3x+1=3x2e3x+1(1+x)f'(x)=u'v+uv'\\=(3x^2\times e^{3x+1})+(x^3\times 3e^{3x+1})\\=3x^2e^{3x+1}+3x^3e^{3x+1}\\=3x^2e^{3x+1}(1+x)

How to Differentiate Logarithmic Functions

To begin our understanding of differentiating logarithmic functions, it is important to start with the reciprocal function. This is a loglog function of base ee denoted by the formula:

ddxlogex=1xorddxlnx=1x\frac{d}{dx}log_ex=\frac{1}{x}\qquad or\qquad \frac{d}{dx}lnx=\frac{1}{x}

Practice Question 2

Differentiate the function, y=7lnxy=7lnx

Applying the formula above we can simply differentiate:

ddx7lnx=7×1x=7x\frac{d}{dx}7lnx=7\times\frac{1}{x}\\=\frac{7}{x}

Now that we have learnt the general formula for the reciprocal function, there are two other standard formulas, which are important for us to apply to various questions.

ddxloge(ax+b)=aax+b\frac{d}{dx}log_e(ax+b)=\frac{a}{ax+b} ddxlogeu=uuorddxlogef(x)=f(x)f(x)\frac{d}{dx}log_eu=\frac{u'}{u}\qquad or\qquad \frac{d}{dx}log_ef(x)=\frac{f'(x)}{f(x)}

By manipulating any expression or equation to fit the form of the formulas above, loglog functions with base ee can be easily solved.

Practice Question 3

Differentiate the function f(x)=loge(x)xf(x)=\frac{log_e(x)}{x} using both the quotient rule, and any related formulas provided above.

We must first determine terms uu and vv to use in the quotient rule:

u=loge(x)v=xu=log_e(x)\qquad v=x

Using differentiation formulas and techniques learnt so far, find uu' and vv':

u=1xv=1u'=\frac{1}{x}\qquad v'=1

Finally, we can substitute these values into the quotient rule formula, to determine the derivative:

f(x)=uvuvv2=1xxloge(x)×1x2=1loge(x)x2f'(x)=\frac{u'v-uv'}{v^2}\\=\frac{\frac{1}{x}x-log_e(x)\times1}{x^2}\\=\frac{1-log_e(x)}{x^2}

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