__In this note, you will learn:__

· __Differentiation with relations to exponential functions__

· __Differentiation with relations to logarithmic functions__

In the past few articles on differentiation, we have learnt about:

Hence, today we are going to be learning about differentiation with relations to exponential and logarithmic functions. Let’s begin!

__Differentiation with relations to exponential functions__

__Differentiation with relations to exponential functions__

If you do not already know, the derivative of an exponential, **e^x**, will yield the same result as an output, which means that:

**d/dx [e^x] = e^x**

However, what if we replace the ‘**x**’ with a **f(x)** instead? What will be the derivative of **e^f(x)**?

Similar to finding the derivative of a trigonometric function, we will have to use the application of Chain Rule to help us out! Let’s take a look at how it’s done:

Let **y = e^u** and **u = f(x)**, hence **dy/du = e^u** and **du/dx = f’(x)**

If we were to apply the Chain Rule here, where dy/dx = (dy/du) x (du/dx),

Hence, we will end up with:

**d/dx [e^f(x)] = e^f(x) x f’(x)**

In a more detailed context, given that f(x) = ax + b, where a and b are constants, then we will obtain the following results by applying the formulae above:

**d/dx [e^ax + b] = a e^ax + b**

Let’s take a look at a few examples of the application of the formulae above:

Differentiate each of the following with respect to x.

a) e^(5 – 4x)

b) 6x^2. e^3x

c) y = e^(3x + 2) sin 3x, where x is in radians

d) f(x) = e^(5x + 3)/ x

Solution:

a) d/dx (e^(5 – 4x)) = e^(5 – 4x) x d/dx (5 – 4x)

Note: To find the derivative of an exponential function, it is basically **the product of the initial exponential function (ef(x))** and **the derivative of the function which the exponential term is being raised to (f(x))**.

In this case, the initial exponential function will be **e^(5 – 4x)** and the derivative of the function which the exponential term is being raised to will be **d/dx (5 – 4x) = 4**.

= e^(5 – 4x) x (- 4)

= - 4 e^(5 – 4x)

b) d/dx (6x^2 . e^3x) = [(6x^2) x d/dx (e^3x)] + [(e^3x) x d/dx (6x^2)]

Note: For this question, we are able to apply the **Product Rule** (**d/dx (uv) = u (dv/dx) x v (du/dx)**) which we have learnt back in the previous articles under differentiation by splitting the terms and letting **u = 6x^2 **and **v = e^3x**, which will result in **du/dx = 12x** and **dv/dx = 3e^3x**.

= (6x^2) (3e^3x) + (e^3x) (12x)

= 18x^2e^3x + 12xe^3x

= 6xe^3x (3x + 2)

c) y = e^3x + 2 sin 3x

Let **u = e^(3x + 2)** and **v = sin 3x**, hence **du/dx = 3e^(3x + 2)** and **dv/dx = 3 cos3x**

dy/dx = (e^(3x + 2)) (3 cos 3x) + (sin 3x) (3 e^(3x + 2))

= 3 cos 3x e^(3x + 2) + 3 sin 3x e^(3x + 2)

= 3 e^(3x + 2) (cos 3x + sin 3x)

d) f(x) = e^(5x + 3)/ x

Note: For this question, we will have to apply the **Quotient Rule** (**d/dx (u/v) = [v(du/dx) – u(dv/dx)]/ v^2**) which we have learnt in the previous articles under differentiation. We will hence have to let u = e^(5x + 3) and v = x, which we will end up with **du/dx = 5 e^(5x + 3)** and **dv/dx = 1**.

f’(x) = [(x) (5 e^(5x + 3)) – (e^(5x + 3)) (1)]/ x^2

= [e^(5x + 3)(5x – 1)]/ x^2

__Differentiation with relations to logarithmic functions__

__Differentiation with relations to logarithmic functions__

Now that we know how to find the derivatives of exponential functions, and we know that logarithm goes hand-in-hand with exponents. That’s right! We will now learn how to find the derivative of logarithmic functions. Let’s begin!

The basic derivative of a logarithmic function is given by:

**d/dx (ln x) = 1/x**

However, what if we replace the ‘**x**’ with a **f(x)** instead? What will be the derivative of **ln f(x)**?

Similar to finding the derivative of an exponential function, we will have to use the application of Chain Rule to help us out! Let’s take a look at how it’s done:

Let **y = ln u** and **u = f(x)**, hence **dy/du = 1/u** and **du/dx = f’(x)**

If we were to apply the Chain Rule here, where dy/dx = (dy/du) x (du/dx),

Hence, we will end up with:

**d/dx [ln f(x)] = (1/u) x f’(x)**

**= f’(x)/ f(x)**

In a more detailed context, given that f(x) = ax + b, where a and b are constants, then we will obtain the following results by applying the formulae above:

**d/dx [ln (ax + b)] = a/(ax + b)**

Let’s take a look at a few examples of the application of the formulae above:

Differentiate each of the following with respect to x.

a) ln (8x^3 + 5x)

b) e^y = 5 cos 2x, where x is in radians.

Solution:

a) d/dx [ln (8x^3 + 5x)] = [1/ (8x^3 + 5x)] x (24x^2 + 5)

Note: The derivative of a logarithmic function is basically **the product of the reciprocal of the function in the argument of the logarithmic term** and **the derivative of the argument of the logarithmic term**.

In this case, the function of the argument of the logarithmic term will be **8x^3 + 5x**, and its reciprocal will be **1/ (8x^3 + 5x)**. The derivative of the argument of the logarithmic term will be **24x^2 + 5**.

= (24x^2 + 5)/ (8x^3 + 5x)

b) e^y = 5 cos 2x

y = ln (5 cos 2x)

Note: For this question, we will have to change the expression from **exponential form** into **logarithmic form**, if you have forgotten how to do so, you can ** check out the articles Math Lobby has for you under the chapter of ‘Logarithm’**!

dy/dx = [1/ (5 cos 2x)] x (-5 sin 2x) (2)

= (-10 sin 2x)/ (5 cos 2x)

= -2 tan 2x

And that’s all for today, students! ** Math Lobby** hopes that after this article, you have a clear understanding on this topic of differentiation with relations to exponential and logarithmic functions, including the application of chain rule for both exponential and logarithmic terms with functions of x!

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