1. Find \( \displaystyle \frac{dy}{dx} \) if \( y = x^2 \) using first principle method.
Solution:
Let \( f(x)=x^2 \) — (i)
Replacing \(x\) with \(x+h\), we get \( f(x+h)=(x+h)^2 \) — (ii)
By definition, \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \)
By (i) & (ii), \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{(x+h)^2 - x^2}{h} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{x^2 + 2xh + h^2 - x^2}{h} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{2xh + h^2}{h} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{h(2x+h)}{h} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}(2x+h) \)
i.e. \( \displaystyle \frac{dy}{dx} = 2x \)
∴ If \( y=x^2 \), we get \( \displaystyle \frac{dy}{dx}=2x \).
Don’t forget
\( \displaystyle \frac{dy}{dx} \)
is a
Gradient Generator.
It gives the gradient of the tangent to the curve at
any point on its graph.
2. Find \( \displaystyle \frac{dy}{dx} \) if \( y = x^3 \) using first principle method.
Solution:
Let \( \displaystyle f(x)=x^3 \) — (i)
Replacing \(x\) with \(x+h\), we get \( \displaystyle f(x+h)=(x+h)^3 \) — (ii)
By definition, \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \)
By (i) & (ii), \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{(x+h)^3 - x^3}{h} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{3x^2h + 3xh^2 + h^3}{h} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{h(3x^2 + 3xh + h^2)}{h} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\big(3x^2 + 3xh + h^2\big) \)
i.e. \( \displaystyle \frac{dy}{dx} = 3x^2 \)
∴ If \( y=x^3 \), we get \( \displaystyle \frac{dy}{dx}=3x^2 \).
If you plug any point on the curve into \( \displaystyle \frac{dy}{dx} \), it tells you whether the graph is increasing, decreasing, or at a turning point.
For example, using the previous problem \( y=x^3 \) and the point \( (2,8) \):
\( \displaystyle \frac{dy}{dx}=3x^2 \quad\Rightarrow\quad \frac{dy}{dx}\Big|_{x=2}=3(2)^2=12 \)
Since \(12\) is +ve, the curve is increasing at \( (2,8) \).
3. Find \( \displaystyle \frac{dy}{dx} \) if \( y=\sin x \) using first principle method.
Solution:
Let \( \displaystyle f(x)=\sin x \) — (i)
Replacing \(x\) with \(x+h\), \( f(x+h)=\sin(x+h) \) — (ii)
By definition, \( \displaystyle \frac{dy}{dx}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \)
By (i) & (ii), \( \displaystyle \frac{dy}{dx}=\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h} \)
∴ \( \displaystyle \frac{dy}{dx} =\lim_{h\to 0}\frac{2\cos\!\big(\frac{x+h+x}{2}\big)\;\sin\!\big(\frac{x+h-x}{2}\big)}{h} \)
i.e. \( \displaystyle \frac{dy}{dx} =\lim_{h\to 0}\frac{2\cos\!\big(\frac{2x+h}{2}\big)\;\sin\!\big(\frac{h}{2}\big)}{h} \)
∴ \( \displaystyle \frac{dy}{dx} =\lim_{h\to 0}\; 2\cos\!\big(\frac{2x+h}{2}\big)\;\frac{\sin(\frac{h}{2})}{h} \)
i.e. \( \displaystyle \frac{dy}{dx} =\lim_{h\to 0}\; \cos\!\big(\frac{2x+h}{2}\big)\;\frac{\sin(\frac{h}{2})}{\frac{h}{2}} \)
i.e. \( \displaystyle \frac{dy}{dx} = \cos\!\big(\frac{2x+0}{2}\big)\times 1 \)
i.e. \( \displaystyle \frac{dy}{dx}=\cos x \)
∴ If \( y=\sin x \), then \( \displaystyle \frac{dy}{dx}=\cos x \).
I guess by now many of you are getting a little annoyed with how long this first-principle method takes 😅
That’s exactly why mathematicians over the last three centuries developed Differentiation Techniques which makes finding gradient of curves faster or let's say smarter ways to find \( \frac{dy}{dx} \).
In the next page, you’ll see shortcuts that skips the "logical" First Principle method and gives the derivatives super fast — methods like the Sum Rule, Product Rule, and Quotient Rule, etc.
4. Find \( \displaystyle \frac{dy}{dx} \) if \( y=\cos x \) using first principle method.
Solution:
Let \( \displaystyle f(x)=\cos x \) — (i)
Replacing \(x\) with \(x+h\), \( \displaystyle f(x+h)=\cos(x+h) \) — (ii)
By definition, \( \displaystyle \frac{dy}{dx}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \)
By (i) & (ii), \( \displaystyle \frac{dy}{dx}=\lim_{h\to 0}\frac{\cos(x+h)-\cos x}{h} \)
∴ \( \displaystyle \frac{dy}{dx} =\lim_{h\to 0}\frac{-2\sin\!\big(\frac{x+h+x}{2}\big)\;\sin\!\big(\frac{x+h-x}{2}\big)}{h} \)
i.e. \( \displaystyle \frac{dy}{dx} =\lim_{h\to 0}\frac{-2\sin\!\big(\frac{2x+h}{2}\big)\;\sin\!\big(\frac{h}{2}\big)}{h} \)
∴ \( \displaystyle \frac{dy}{dx} =\lim_{h\to 0}\; -\sin\!\big(\frac{2x+h}{2}\big)\;\frac{\sin(\frac{h}{2})}{\frac{h}{2}} \)
i.e. \( \displaystyle \frac{dy}{dx} = -\sin\!\big(\frac{2x+0}{2}\big)\times 1 \)
i.e. \( \displaystyle \frac{dy}{dx}=-\sin x \)
∴ If \( y=\cos x \), then \( \displaystyle \frac{dy}{dx}=-\sin x \).
Now if you want to try some challenging problems in First Principles, you can go to the next questions.
If not, you can hit the Next Topic button and start learning Differentiation Techniques — faster and smarter ways to find \( \displaystyle \frac{dy}{dx} \).
5. Find \( \displaystyle \frac{dy}{dx} \) if \( y=\tan x \) using first principle method.
Solution:
Let \( f(x)=\tan x \) — (i)
Replacing \(x\) with \(x+h\), \( f(x+h)=\tan(x+h) \) — (ii)
By definition, \( \displaystyle \frac{dy}{dx}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \)
By (i) & (ii), \( \displaystyle \frac{dy}{dx}=\lim_{h\to 0}\frac{\tan(x+h)-\tan x}{h} \)
∴ \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{\sin((x+h)-x)}{h\cos(x+h)\cos x} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{\sin h}{h\cos(x+h)\cos x} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{\sin h}{h} \times \frac{1}{\cos(x+h)\cos x} \)
i.e. \( \displaystyle \frac{dy}{dx} = 1 \times \frac{1}{\cos^2 x} \)
i.e. \( \displaystyle \frac{dy}{dx} = \sec^2 x \)
∴ If \( y=\tan x \), then \( \displaystyle \frac{dy}{dx}=\sec^2 x \).
READY FOR THE NEXT CHALLENGE?
Try it yourself: find the derivative of \( y=\cot x \) using first principles. NO CHEATING!!
The solution is below — work it out first and then check your steps.
6. Find \( \displaystyle \frac{dy}{dx} \) if \( y=\cot x \) using first principle method.
Solution:
Let \( f(x)=\cot x \) — (i)
Replacing \(x\) with \(x+h\), \( f(x+h)=\cot(x+h) \) — (ii)
By definition, \( \displaystyle \frac{dy}{dx}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \)
By (i) & (ii), \( \displaystyle \frac{dy}{dx}=\lim_{h\to 0}\frac{\cot(x+h)-\cot x}{h} \)
∴ \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{\sin\!\big(x-(x+h)\big)}{h\,\sin(x+h)\,\sin x} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{\sin(-h)}{h\,\sin(x+h)\,\sin x} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\frac{-\sin h}{h\,\sin(x+h)\,\sin x} \)
i.e. \( \displaystyle \frac{dy}{dx} = \lim_{h\to 0}\left(-\,\frac{\sin h}{h}\right)\times\frac{1}{\sin(x+h)\,\sin x} \)
i.e. \( \displaystyle \frac{dy}{dx} = -1 \times \frac{1}{\sin^2 x} \)
i.e. \( \displaystyle \frac{dy}{dx} = -\,\csc^{2} x \)
∴ If \( y=\cot x \), then \( \displaystyle \frac{dy}{dx}=-\,\csc^{2} x \).
Do you know?
American textbooks usually write csc x, while British/Commonwealth texts often write cosec x. They mean the same thing: \( \displaystyle \csc x=\text{cosec }x=\frac{1}{\sin x} \).
READY FOR PROBLEM 7?
Next up: find the derivative of \( y=\sec x \) using first principles.