Differentiation of Exponential and Logarithmic Functions

Expert reviewed 08 January 2025 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:

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How to Differentiate Exponential Functions

Note:

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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

Note:

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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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