Problems on Algebra of Derivative
Problem 1: Find the derivative for the given function f(x).
f(x) = x2 + 3x
Solution:
This function is the sum of two different function. Sum rule will be used here.
f(x) = x2 + 3x
Here, h(x) = x2 and g(x) = 3x.
f(x) = h(x) + g(x)
⇒f'(x) = h'(x) + g'(x)
⇒ f'(x) = [Tex]\frac{d}{dx}h(x) + \frac{d}{dx}g(x) [/Tex]
⇒f'(x) =[Tex]\frac{d}{dx}x^2 + \frac{d}{dx}3x [/Tex]
⇒f'(x) = 2x + 3
Problem 2: Find the derivative for the given function f(x).
f(x) = ex + sin(x)
Solution:
This function is the sum of two different function. Sum rule will be used here.
f(x) = ex + sin(x)
Here, h(x) =ex and g(x) = sin(x)
f(x) = h(x) + g(x)
⇒f'(x) = h'(x) + g'(x)
⇒ f'(x) = [Tex]\frac{d}{dx}h(x) + \frac{d}{dx}g(x) [/Tex]
⇒f'(x) =[Tex]\frac{d}{dx}e^x + \frac{d}{dx}sin(x) [/Tex]
⇒f'(x) = ex + cos(x)
Problem 3: Find the derivative for the given function f(x),
f(x) = 5x4 – 3x2
Solution:
This function is the difference of two different function. Difference rule will be used here.
f(x) = 5x4 – 3x2
Here, h(x) =5x4 and g(x) = 3x2
f(x) = h(x) – g(x)
⇒f'(x) = h'(x) – g'(x)
⇒ f'(x) = [Tex]\frac{d}{dx}h(x) – \frac{d}{dx}g(x) [/Tex]
⇒f'(x) =[Tex]\frac{d}{dx}5x^4 – \frac{d}{dx}3x^2 [/Tex]
⇒f'(x) = 20x3 – 6x
Problem 4: Find the derivative for the given function f(x),
f(x) = 5log(x) – 3x
Solution:
This function is the difference of two different function. Difference rule will be used here.
f(x) = 5log(x) – 3x
Here, h(x) =5log(x) and g(x) = 3x
f(x) = h(x) – g(x)
⇒f'(x) = h'(x) – g'(x)
⇒ f'(x) = [Tex]\frac{d}{dx}h(x) – \frac{d}{dx}g(x) [/Tex]
⇒f'(x) =[Tex]\frac{d}{dx}5log(x) – \frac{d}{dx}3x [/Tex]
⇒f'(x) = [Tex]\frac{5}{x} – 3 [/Tex]
Problem 5: Find the derivative for the given function f(x),
f(x) = 5x4.sin(x)
Solution:
This function is the product of two different function. Product rule will be used here.
f(x) = 5x4.sin(x)
Here, h(x) =5x4 and g(x) = sin(x)
f(x) = h(x).g(x)
⇒f'(x) = h'(x) g(x) + h(x)g'(x)
⇒ f'(x) = [Tex]\frac{d}{dx}(f(x))g(x) + \frac{d}{dx}(g(x))f(x) [/Tex]
⇒f'(x) =[Tex]\frac{d}{dx}(5x^4)sin(x) + \frac{d}{dx}(sin(x))5x^4 [/Tex]
⇒f'(x) = 20x3sin(x) + 5x4cos(x)
Problem 6: Find the derivative for the given function f(x),
f(x) = 5ex.log(x)
Solution:
This function is the product of two different function. Product rule will be used here.
f(x) = 5ex.log(x)
Here, h(x) =5ex and g(x) = log(x)
f(x) = h(x).g(x)
⇒f'(x) = h'(x) g(x) + h(x)g'(x)
⇒ f'(x) = [Tex]\frac{d}{dx}(f(x))g(x) + \frac{d}{dx}(g(x))f(x) [/Tex]
⇒f'(x) =[Tex]\frac{d}{dx}(5e^x)log(x) + \frac{d}{dx}(log(x))5e^x [/Tex]
⇒f'(x) = [Tex]5(e^xlog(x) + \frac{1}{x}e^x) [/Tex]
Problem 7: Find the derivative for the given function f(x),
f(x) = [Tex]\frac{x + 1}{2x} [/Tex]
Solution:
This function is the division of two different function. Division rule will be used here.
f(x) =[Tex]\frac{x + 1}{2x} [/Tex]
Here, h(x) =x + 1 and g(x) = 2x
f(x) = [Tex]\frac{h(x)}{g(x)} [/Tex]
⇒f'(x) = [Tex]\frac{g(x)h'(x) – h(x)g'(x)}{(g(x))^2} [/Tex]
⇒ f'(x) = [Tex]\frac{\frac{d}{dx}(h(x))g(x) – \frac{d}{dx}(g(x))h(x)}{(g(x))^2} [/Tex]
⇒f'(x) =[Tex]\frac{\frac{d}{dx}(x + 1)2x – \frac{d}{dx}(2x)(x + 1)}{(x + 1)^2} [/Tex]
⇒f'(x) =[Tex]\frac{2x – 2(x + 1)}{(x + 1)^2} [/Tex]
⇒f'(x) = [Tex]\frac{-2}{(x + 1)^2} [/Tex]
Problem 8: Find the derivative for the given function f(x),
f(x) = [Tex]\frac{log(x)}{2x} [/Tex]
Solution:
This function is the division of two different function. Division rule will be used here.
f(x) =[Tex]\frac{log(x)}{2x} [/Tex]
Here, h(x) =log(x) and g(x) = 2x
f(x) = [Tex]\frac{h(x)}{g(x)} [/Tex]
⇒f'(x) = [Tex]\frac{g(x)h'(x) – h(x)g'(x)}{(g(x))^2} [/Tex]
⇒ f'(x) = [Tex]\frac{\frac{d}{dx}(h(x))g(x) – \frac{d}{dx}(g(x))h(x)}{(g(x))^2} [/Tex]
⇒f'(x) =[Tex]\frac{\frac{d}{dx}(log(x))2x – \frac{d}{dx}(2x)(log(x))}{4x^2} [/Tex]
⇒f'(x) =[Tex]\frac{1- 2log(x)}{4x^2} [/Tex]
Algebra of Derivative of Functions
Derivatives are an integral part of calculus. They measure the rate of change in any quantity. Suppose there is a water tank from which water is leaking. A local engineer is asked to measure the time in which the water tank will become empty. In such a scenario, the engineer needs to know two things — the size of the water tank and the rate at which water is flowing out of it. The size of the tank can be found out easily but to measure the rate at which water is leaking he will have to use derivatives.
In this way, derivatives are intertwined in our lives. It is easy to calculate the derivatives for simple functions, but when functions become complex the correct way to approach this problem is to break the problem into subproblems that are easier to solve. Let’s see some rules and approaches to do that in the case of derivatives.
Table of Content
- What is Derivatives?
- Rules of Differentiation
- Problems on Algebra of Derivatives
- FAQs