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

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

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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