Differentiation

Lesson

Just like we saw in derivatives of a sum of the type x^n,
#### Examples

#### Worked Examples:

##### question 1

##### question 2

##### question 3

So we can see that the derivative of the sum is the same as the sums of the derivatives.

Sum of Derivatives

Derivative of sum is equal to the sum of the derivatives.

If $f(x)=g(x)\pm h(x)$`f`(`x`)=`g`(`x`)±`h`(`x`) then $f'(x)=g'(x)\pm h'(x)$`f`′(`x`)=`g`′(`x`)±`h`′(`x`)

This means we can apply the power rule to individual terms.

And this applies to any function, whether it be the $x^n$`x``n` we saw before, or $ax^n$`a``x``n` that we are exploring now.

Find the derivative of the following,

a) $f(x)=4x^2+3x+2$`f`(`x`)=4`x`2+3`x`+2, then $f'(x)=8x+3$`f`′(`x`)=8`x`+3 (remember that the derivative of a constant term is $0$0)

b) $f(x)=3x^3-3x^2$`f`(`x`)=3`x`3−3`x`2, then $f'(x)=9x^2-6x$`f`′(`x`)=9`x`2−6`x`

c) $f(x)=6x^{-3}-2x+\sqrt{x}$`f`(`x`)=6`x`−3−2`x`+√`x`. Firstly we need to turn the $\sqrt{x}$√`x` into a power. $\sqrt{x}=x^{\frac{1}{2}}$√`x`=`x`12 So $f(x)=6x^{-3}-2x+x^{\frac{1}{2}}$`f`(`x`)=6`x`−3−2`x`+`x`12 and so then the derivative $f'(x)=-18x^{-4}-2+\frac{1}{2}x^{-\frac{1}{2}}$`f`′(`x`)=−18`x`−4−2+12`x`−12

Differentiate $y=7x^2-9x+8$`y`=7`x`2−9`x`+8.

Differentiate $y=\frac{24}{x^5}-\frac{30}{x^4}$`y`=24`x`5−30`x`4.

Find the derivative of $y=x^3\sqrt{x}+3x^5$`y`=`x`3√`x`+3`x`5.

Apply differentiation and anti-differentiation techniques to polynomials

Apply calculus methods in solving problems