So we can see that the derivative of the sum is the same as the sums of the 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 $x^n$xn we saw before, or $ax^n$axn that we are exploring now.
Find the derivative of the following,
a) $f(x)=4x^2+3x+2$f(x)=4x2+3x+2, then $f'(x)=8x+3$f′(x)=8x+3 (remember that the derivative of a constant term is $0$0)
b) $f(x)=3x^3-3x^2$f(x)=3x3−3x2, then $f'(x)=9x^2-6x$f′(x)=9x2−6x
c) $f(x)=6x^{-3}-2x+\sqrt{x}$f(x)=6x−3−2x+√x. Firstly we need to turn the $\sqrt{x}$√x into a power. $\sqrt{x}=x^{\frac{1}{2}}$√x=x12 So $f(x)=6x^{-3}-2x+x^{\frac{1}{2}}$f(x)=6x−3−2x+x12 and so then the derivative $f'(x)=-18x^{-4}-2+\frac{1}{2}x^{-\frac{1}{2}}$f′(x)=−18x−4−2+12x−12
Differentiate $y=7x^2-9x+8$y=7x2−9x+8.
Differentiate $y=\frac{24}{x^5}-\frac{30}{x^4}$y=24x5−30x4.
Find the derivative of $y=x^3\sqrt{x}+3x^5$y=x3√x+3x5.