Sum and Difference Rule of Derivatives

Sum and difference rule of derivative

Theorem:

Let f and g are differentiable at x ,

Then (f+g) and (f-g) are also differentiable at x and

\left[f\left(x\right)+g\left(x\right)\right]'=f'\left(x\right)+g'\left(x\right)

That is

\frac d{dx}\left[f\left(x\right)+g\left(x\right)\right]=\frac d{dx}\left[f\left(x\right)\right]+\frac d{dx}\left[g\left(x\right)\right]

also

\left[f\left(x\right)-g\left(x\right)\right]'=f'\left(x\right)-g'\left(x\right)
Sum and Difference Rule of Derivatives

That is

\frac d{dx}\left[f\left(x\right)-g\left(x\right)\right]=\frac d{dx}\left[f\left(x\right)\right]-\frac d{dx}\left[g\left(x\right)\right]

Proof
You can prove sum and difference derivatives in easy way by using following pattren.
Step 1:    Let function.........eq(1)
Step 2:    adding \delta y and \delta x in y and x component respectively .......eq(2)
Step 3:  Subtracting eq(2)-eq(1)
Step 4:  Dividing \delta x post cycle therapy pct on both sides.
Step 5:  applying \underset{\delta x\rightarrow0}{lim} on both sides.
Step 1:

Let ϕ(x)=f(x)+g(x)..............(1)

Step 2:    Adding \delta x on both sides

\phi(x+\delta x)=f(x+\delta x)+g(x+\delta x)..........(2)

Step 3: Subtracting eq(2)-eq(1)

\phi(x+\delta x)-\phi(x)

=\left[f(x+\delta x))+g(x+\delta x)\right]-\left[f(x)+g(x)\right]

\phi(x+\delta x)-\phi(x)=f(x+\delta x)+g(x+\delta x)-f(x)-g(x)

\phi(x+\delta x)-\phi(x)=[f(x+\delta x)-f(x)]+[g(x+\delta x)-g(x)]

Step 4: Dividing \delta x on both sides

\frac{\phi(x+\delta x)-\phi(x)}{\delta x}=\frac{\lbrack f(x+\delta x)-f(x)\rbrack}{\delta x}+\frac{\lbrack g(x+\delta x)-g(x)\rbrack}{\delta x}

Step 5: Taking \underset{\delta x\rightarrow0}{lim} on both sides

\underset{\delta x\rightarrow0}{lim}\frac{\phi(x+\delta x)-\phi(x)}{\delta x}=\underset{\delta x\rightarrow0}{lim}\left[\frac{\lbrack f(x+\delta x)-f(x)\rbrack}{\delta x}+\frac{\lbrack g(x+\delta x)-g(x)\rbrack}{\delta x}\right]

\underset{\delta x\rightarrow0}{lim}\frac{\phi(x+\delta x)-\phi(x)}{\delta x}=\underset{\delta x\rightarrow0}{lim}\left[\frac{\lbrack f(x+\delta x)-f(x)\rbrack}{\delta x}\right]+\underset{\delta x\rightarrow0}{lim}\left[\frac{\lbrack g(x+\delta x)-g(x)\rbrack}{\delta x}\right]

\phi'(x)=f'(x)+g'(x)

\frac d{dx}\phi(x)=\frac d{dx}f(x)+\frac d{dx}g(x)

Example 1:
Sum and Difference Rule of Derivatives

Sum and difference rule of derivatives

f(x)=3x^5 and g(x)=4x

Sum of derivatives 

\frac d{dx}\left[f(x)+g(x)\right]=\frac d{dx}\left[3x^5\right]+\frac d{dx}\left[4x\right]

\frac d{dx}\left[f(x)+g(x)\right]=3\left(5\right)x^{5-1}+4\left(1\right)

\frac d{dx}\left[f(x)+g(x)\right]=15x^4+4

difference of derivatives 

\frac d{dx}\left[f(-g(x)\right]=\frac d{dx}\left[3x^5\right]x)-\frac d{dx}\left[4x\right]

\frac d{dx}\left[f(x)-g(x)\right]=3\left(5\right)x^{5-1}-4\left(1\right)

\frac d{dx}\left[f(x)-g(x)\right]=15x^4-4

Example 2:
Sum and Difference Rule of Derivative

sum and difference rule of derivative

f(x)=4x^3 and g(x)=3x

Sum of derivatives 

\frac d{dx}\left[f(x)+g(x)\right]=\frac d{dx}\left[4x^3\right]+\frac d{dx}\left[3x\right]

\frac d{dx}\left[f(x)+g(x)\right]=4\left(3\right)x^2+3\left(1\right)

\frac d{dx}\left[f(x)+g(x)\right]=12x^2+3

\frac d{dx}\left[f(x)+g(x)\right]=3\left[4x^2+1\right]

difference of derivatives 

\frac d{dx}\left[f(x)-g(x)\right]=\frac d{dx}\left[4x^3\right]-\frac d{dx}\left[3x\right]

\frac d{dx}\left[f(x)-g(x)\right]=4\left(3\right)x^2-3\left(1\right)

\frac d{dx}\left[f(x)-g(x)\right]=12x^2-3

\frac d{dx}\left[f(x)-g(x)\right]=3\left[4x^2-1\right]

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Azhar Ali

Azhar Ali

I graduated in Mathematics from the University of Sargodha, having master degree in Mathematics.

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