Derivatives of Sums, Differences, and Constant Multiples

Ok that was too easy. :)

Ok, no more fooling around.

Huh. That's pretty convenient. Do you see what happened?

Well, the derivative of $e^x + \sin (x)$ turned out just to be the derivative of $e^x$ plus the derivative of $\sin (x).$

That's exactly why we have math notation! In general, we can say,

If $f(x)$ and $g(x)$ are differentiable functions, then $(f(x)+g(x))' = f'(x) + g'(x).$

Note that in the definition, $f$ and $g$ obviously need to be differentiable at $x,$ or else we can't take the derivative! We formally state this as

If $f(x)$ and $g(x)$ are differentiable functions, then so is $f(x)+g(x).$

Ok, what about this.

Great! We can formalize this like the "additive rule" too.

If $f(x)$ is a differentiable function, then for all real numbers $a,$ we have $a\cdot f(x)$ is differentiable and $(a\cdot f(x))' = a\cdot f'(x).$

These two rules give us that derivatives are linear, or we can also say that this property is called Linearity of Derivatives.

If $f(x)$ and $g(x)$ are differentiable functions, then $(f(x)+g(x))' = f'(x) + g'(x).$

If $f(x)$ is a function, then for all real numbers $a,$ we have $(a\cdot f(x))' = a\cdot f'(x).$

You might also want to check out other topics on linearity. You'll notice this is a fairly ubiquitous concept :)

Finally, we can combine the two properties: If $f(x)$ and $g(x)$ are functions, then for all real numbers $a$ and $b,$ we have $(a\cdot f(x) + b\cdot g(x))' = a\cdot f'(x) + b\cdot g'(x)$, provided each of the last two derivatives exists.

Linearity of Definite Integrals

Linearity of Limits (sequences)

PROOF OF LINEARITY OF DERIVATIVE

First, we know that a limit can be expressed (decomposed) as the sum of limits given that each individual limit exists. In other words, $\lim f(x) + \lim g(x) = \lim [f(x) + g(x)]$. The proof of this requires a careful argument of $\epsilon - \delta$.

Because a derivative is by definition a limit, we can use the limit sum formula to deduce that $f'(x) + g'(x) = (f(x) + g(x))'$ if and only if the individual derivative of each function exists.

Note that, also by $\epsilon - \delta$, $\lim [af(x)] = a \lim [f(x)]$. A similar argument can prove that $[af(x)]' = af'(x)$, considering that $af(x+b) - af(x) = a(f(x+b) - f(x)).$

Thus, the derivative is linear.

The linearity of derivatives refers to how they can be added or subtracted from each other. We state that

There is a proof using the limit definition of the derivative.

Let $j(x) = f(x) + g(x)$. Then by the limit definition of the derivative:

$$j'(x) = \displaystyle\lim_{h \rightarrow 0}\frac{j(x + h) - j(x)}{h}$$

However, as $j(x) = f(x) + g(x)$, the right-hand side can be written as

$$j'(x) = \displaystyle\lim_{h \rightarrow 0}\frac{f(x + h) + g(x + h) - f(x) - g(x)}{h}$$

Now the limit can be broken into two:

$$j'(x) = \displaystyle\lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h} + \lim_{h \rightarrow 0}\frac{g(x + h) - g(x)}{h}$$

By the limit definition of the derivative, the two limits on the right-hand side can be substituted.

$j'(x) = f'(x) + g'(x) \Rightarrow$ $(j(x))' = f'(x) + g'(x) \Rightarrow$ $(f(x) + g(x))' = f'(x) + g'(x)$

This tool is particularly practical for differentiating polynomials, which are sums of monomials that can be differentiated individually. Now these derivatives can be added together.

(Video) Solving Every Derivative Ever

by PatrickJMT

Okay, maybe it's not EVERY derivative ever but here's a video by patrickJMT going over different derivatives for a whopping 35 minutes.

Let's have a season recap of some formulas before we go in. $$\begin{alignat}{2} & \frac{d}{dx}x^n=nx^{n-1} \quad && \frac{d}{dx}\cot x=-\csc^2 x\\ & \frac{d}{dx}k=0 \quad && \frac{d}{dx}\sec x=\sec x \tan x\\ & \frac{d}{dx}\sin x=\cos x \quad && \frac{d}{dx}e^x=e^x\\ & \frac{d}{dx}\cos x=-\sin x \quad && \frac{d}{dx}a^x=a^x\ln a\\ &\frac{d}{dx}\tan x=\sec^2x \quad && \frac{d}{dx}\ln x=\frac1x\\ &\frac{d}{dx}\csc x=-\csc x\cot x \quad &&\frac{d}{dx}\log_ax=\frac1{x\ln a} \end{alignat}$$

Also let's recap the product, quotient, and chain rule. $$\begin{align} [f(x)g(x)]'&=f'(x)g(x)+f(x)g'(x)\\ \bigg[\frac{f(x)}{g(x)}\bigg]'&=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\\ [f(\ g(x)\ )]'&=f'(\ g(x)\ )\cdot g'(x) \end{align}$$

Problem 1

$$\begin{align} y&=x^2+3x+4\\ y'&=2x^{2-1}+3(1x^{1-1})+0\\ &=\boxed{2x+3} \end{align}$$

Problem 2

$$\begin{align} y&=5t^{-\frac23}\\ y'&=5(-\frac23t^{-\frac23-1})\\ &=-\frac{10}3t^{-\frac53}\\ &=\boxed{-\frac{10}{3t^{\frac53}}} \end{align}$$

Problem 3

$$\begin{align} y&=\frac{3x-1}{\sqrt x}\\ &=\frac{3x-1}{x^{\frac12}}\\ &=\frac{3x}{x^{\frac12}}-\frac{1}{x^{\frac12}}\\ &=3x^{\frac12}-x^{-\frac12}\\ y'&=3(\frac12x^{-\frac12})-(-\frac12x^{-\frac32})\\ &=\frac{3}{2x^{\frac12}}+\frac{1}{2x^{\frac32}}\\ &=\boxed{\frac{3x+1}{2x^{\frac32}}} \end{align}$$

Problem 4

$$\begin{align} y&=(x+4)(x-4)\\ &=x^2-4x+4x-16\\ &=x^2-16\\ y'&=2x^{2-1}-0\\ &=\boxed{2x} \end{align}$$

Problem 5

$$\begin{align} y&=\frac{\sqrt x-1}{\sqrt x+1}\\ &=\frac{x^{\frac12}-1}{x^{\frac12}+1}\\ y'&=\frac{\color{red}{(x^{\frac12}-1)'}(x^{\frac12}+1)-(x^{\frac12}-1)\color{red}{(x^{\frac12}+1)'}}{(x^{\frac12}+1)^2}\\ &=\frac{\frac12x^{-\frac12}(x^{\frac12}+1)-(x^{\frac12}-1)(\frac12x^{-\frac12})}{(x^{\frac12}+1)^2}\\ &=\frac{\frac12x^{-\frac12}[(x^{\frac12}+1)-(x^{\frac12}-1)]}{(x^{\frac12}+1)^2}\\ &=\frac{(x^{\frac12}+1)-(x^{\frac12}-1)}{2\sqrt x(x^{\frac12}+1)^2}\\ &=\frac{2}{2\sqrt x(x^{\frac12}+1)^2}\\ &=\boxed{\frac{1}{\sqrt x(x^{\frac12}+1)^2}} \end{align}$$

Problem 6

$$\begin{align} y&=\frac{2x-\sqrt{x}\cdot x}{x^2}\\ &=\frac{2x-x^{\frac32}}{x^2}\\ &=\frac{2x}{x^2}-\frac{x^{\frac32}}{x^2}\\ &=\frac{2}{x}-\frac{1}{x^{\frac12}}\\ &=2x^{-1}-x^{-\frac12}\\ y'&=2(-1x^{-2})-(-\frac12x^{\frac32})\\ &=\boxed{\frac{1}{2x^{\frac32}}-\frac{2}{x^2}} \end{align}$$

Problem 7

$$\begin{align} y&=(\sin x)(x^{\frac12})\\ y'&=\color{red}{(\sin x)'}(x^{\frac12})+(\sin x)\color{red}{(x^{\frac12})'}\\ &=(\cos x)(x^{\frac12})+(\sin x)(\frac12x^{-\frac12})\\ &=\boxed{\sqrt{x}\cos x+\frac{\sin x}{2\sqrt x}} \end{align}$$

Problem 8

$$\begin{align} y&=(\sin x)(\cos x)\\ y'&=\color{red}{(\sin x)'}(\cos x)+(\sin x)\color{red}{(\cos x)}\\ &=(\cos x)(\cos x)+(\sin x)(-\sin x)\\ &=\cos^2x-\sin^2x\\ &=\boxed{\cos(2x)} \end{align}$$

Problem 9

$$\begin{align} y&=\sin(\cos x)\\ y'&=\cos(\cos x)\cdot\color{red}{[\cos x]'}\\ &=\cos(\cos x)\cdot(-\sin x)\\ &=\boxed{-\sin x\cos(\cos x)} \end{align}$$

Problem 10

$$\begin{align} y&=\sec[(1+x^2)^{\frac12}]\\ y'&=\sec[(1+x^2)^{\frac12}]\cdot\tan[(1+x^2)^{\frac12}]\cdot\color{red}{[(1+x^2)^{\frac12}]'}\\ &=\sec[(1+x^2)^{\frac12}]\cdot\tan[(1+x^2)^{\frac12}]\cdot\frac12(1+x^2)^{-\frac12}\cdot\color{red}{(1+x^2)'}\\ &=\sec[(1+x^2)^{\frac12}]\cdot\tan[(1+x^2)^{\frac12}]\cdot\frac12(1+x^2)^{-\frac12}\cdot(2x)\\ &=\boxed{\frac{x\sec[(1+x^2)^{\frac12}]\tan[(1+x^2)^{\frac12}]}{\sqrt{1+x^2}}} \end{align}$$

Problem 11

$$\begin{align} y&=\tan^2[\sin(\theta^3)]\\ y'&=2\tan[\sin(\theta^3)]\cdot\color{red}{[\ \tan[\sin(\theta^3)]\ ]'}\\ &=2\tan[\sin(\theta^3)]\cdot\sec^2[\sin(\theta^3)]\cdot\color{red}{[\sin(\theta^3)]'}\\ &=2\tan[\sin(\theta^3)]\cdot\sec^2[\sin(\theta^3)]\cdot\cos(\theta^3)\cdot\color{red}{(\theta^3)'}\\ &=2\tan[\sin(\theta^3)]\cdot\sec^2[\sin(\theta^3)]\cdot\cos(\theta^3)\cdot3\theta^2\\ &=\boxed{6\theta^2\tan[\sin(\theta^3)]\sec^2[\sin(\theta^3)]\cos(\theta^3)} \end{align}$$

Problem 12

$$\begin{align} y&=x\sin(3x^{-1})\\ y'&=\color{red}{(x)'}[\sin(3x^{-1})]+(x)\color{red}{[\sin(3x^{-1})]'}\\ &=(1)\sin(3x^{-1})+(x)\cos(3x^{-1})\cdot\color{red}{(3x^{-1})'}\\ &=\sin(3x^{-1})+(x)\cos(3x^{-1})\cdot3(-1x^{-2})\\\\ &=\sin(3x^{-1})-\frac{3x\cos(3x^{-1})}{x^{2}}\\ &=\sin(\frac3x)-\frac{3\cos(\frac3x)}{x}\\ &=\boxed{\frac{x\sin(\frac3x)-3\cos(\frac3x)}{x}} \end{align}$$

Problem 13

$$\begin{align} y&=\frac{(x^3+4)^2}{(x^2+1)^4}\\ y'&=\frac{\color{red}{[(x^3+4)^2]'}(x^2+1)^4-(x^3+4)^2\color{red}{[(x^2+1)^4]'}}{[(x^2+1)^4]^2}\\ &=\frac{2(x^3+4)\color{red}{(x^3+4)'}(x^2+1)^4-4(x^3+4)^2(x^2+1)^3\color{red}{(x^2+1)'}}{(x^2+1)^8}\\ &=\frac{6x^2(x^3+4)(x^2+1)^4-8x(x^3+4)^2(x^2+1)^3}{(x^2+1)^8}\\ &=\frac{6x^2(x^3+4)(x^2+1)^4-8x(x^3+4)^2(x^2+1)^3}{(x^2+1)^8}\\ &=\frac{2x(x^3+4)(x^2+1)^3[3x(x^2+1)-4(x^3+4)]}{(x^2+1)^8}\\ &=\frac{2x(x^3+4)[3x(x^2+1)-4(x^3+4)]}{(x^2+1)^5}\\ &=\frac{2x(x^3+4)[(3x^3+3x)-(4x^3+16)]}{(x^2+1)^5}\\ &=\boxed{\frac{2x(x^3+4)(-x^3+3x-16)}{(x^2+1)^5}} \end{align}$$

Problem 14

$$\begin{align} y&=\ln(x)+x^3\\ y'&=\frac1x+3x^2\\ &=\boxed{\frac{3x^3+1}{x}} \end{align}$$

Problem 15

$$\begin{align} y&=x^2e^x\\ y'&=\color{red}{(x^2)'}e^x+x^2\color{red}{(e^x)'}\\ &=2xe^x+x^2e^x\\ &=\boxed{xe^x(x+2)} \end{align}$$

Problem 16

$$\begin{align} y&=e^{t\cdot\cos(2t)}\\ y'&=e^{t\cdot\cos(2t)}\cdot\color{red}{[t\cdot\cos(2t)]'}\\ &=e^{t\cdot\cos(2t)}\cdot[\color{red}{(t)'}\cos(2t)+t\color{red}{[\cos(2t)]'}]\\ &=e^{t\cdot\cos(2t)}\cdot[\cos(2t)-t\sin(2t)\cdot\color{red}{(2t)'}]\\ &=\boxed{e^{t\cdot\cos(2t)}\cdot[\cos(2t)-2t\sin(2t)]} \end{align}$$

Problem 17

$$\begin{align} y&=\cos(e^{\pi x})\\ y'&=-\sin(e^{\pi x})\cdot\color{red}{(e^{\pi x})'}\\ &=-\sin(e^{\pi x})\cdot e^{\pi x}\color{red}{(\pi x)'}\\ &=\boxed{-\pi e^{\pi x}\sin(e^{\pi x})} \end{align}$$

Problem 18

$$\begin{align} y&=(1+2e^{3x})^{\frac12}\\ y'&=\frac12(1+2e^{3x})^{-\frac12}\cdot\color{red}{(1+2e^{3x})'}\\ &=\frac12(1+2e^{3x})^{-\frac12}\cdot2e^{3x}\cdot\color{red}{(3x)'}\\ &=\boxed{\frac{3e^{3x}}{\sqrt{1+2e^{3x}}}} \end{align}$$

Problem 19

$$\begin{align} y&=5^{-\frac1x}\\ &=5^{-x^{-1}}\\ y'&=5^{-x^{-1}}\cdot\ln5\cdot\color{red}{(-x^{-1})'}\\ &=5^{-x^{-1}}\cdot\ln5\cdot(x^{-2})\\ &=\boxed{\frac{\ln5}{5^{\frac1x}x^2}} \end{align}$$

Problem 20

$$\begin{align} y&=\ln(x^4\sin^2x)\\ y'&=\frac{1}{x^4\sin^2x}\cdot\color{red}{[x^4\sin^2x]'}\\ &=\frac{1}{x^4\sin^2x}\cdot[\color{red}{(x^4)'}\sin^2x+x^4\color{red}{(\sin^2x)'}]\\ &=\frac{4x^3\sin^2x+2x^4\sin x\cdot\color{red}{(\sin x)'}}{x^4\sin^2x}\\ &=\frac{4x^3\sin^2x+2x^4\sin x\cos x}{x^4\sin^2x}\\ &=\frac{2x^3\sin x(2\sin x+x\cos x)}{x^4\sin^2x}\\ &=\frac{2(2+x\cos x)}{x\sin x}\\ &=\boxed{\frac4x+2\cot x} \end{align}$$

Problem 21

$$\begin{align} y&=\ln\sqrt{\frac{5x-1}{2x+3}}\\ &=\ln\bigg(\frac{5x-1}{2x+3}\bigg)^{\frac12}\\ &=\frac12\bigg[\ln\bigg(\frac{5x-1}{2x+3}\bigg)\bigg]\\ &=\frac12[\ln(5x-1)-\ln(2x+3)]\\ y'&=\frac12[\frac{1}{5x-1}\cdot\color{red}{(5x-1)'}-\frac{1}{2x+3}\cdot\color{red}{(2x+3)'}]\\ &=\frac12\bigg[\frac{5}{5x-1}-\frac{2}{2x+3}\bigg]\\ &=\frac12\bigg[\frac{5(2x+3)-2(5x-1)}{(5x-1)(2x+3)}\bigg]\\ &=\frac12\bigg[\frac{10x+15--10x+2}{10x^2+13x-3}\bigg]\\ &=\frac12\bigg[\frac{17}{10x^2+13x-3}\bigg]\\ &=\boxed{\frac{17}{20x^2+26x-6}} \end{align}$$

Problem 22

$$\begin{align} y&=t^3-3^t\\ y'&=3t^2-3^t\cdot\ln3\\ &=\boxed{3t^2-3^t\ln3} \end{align}$$

Problem 23

$$\begin{align} y&=\ln(\sec x+\tan x)\\ y'&=\frac{1}{\sec x+\tan x}\cdot\color{red}{[\sec x+\tan x]'}\\ &=\frac{1}{\sec x+\tan x}\cdot[\sec x\cdot\tan x+\sec^2x]\\ &=\boxed{\frac{\sec x\tan x+\sec^2x}{\sec x+\tan x}} \end{align}$$

Problem 24

$$\begin{align} y&=\log_{10}\bigg(\frac{2x}{x-1}\bigg)\\ &=\log_{10}(2x)-\log_{10}(x-1)\\ y'&=\frac{1}{2x\ln10}\cdot\color{red}{(2x)'}-\frac{1}{(x-1)\ln10}\cdot\color{red}{(x-1)'}\\ &=\frac{2}{2x\ln10}-\frac{1}{(x-1)\ln10}\\ &=\boxed{\frac{1}{x\ln10}-\frac{1}{(x-1)\ln10}} \end{align}$$

Problem 25

$$\begin{align} y&=\frac{\ln x}{x^2}\\ y'&=\frac{\color{red}{(\ln x)'}(x^2)-(\ln x)\color{red}{(x^2)'}}{[x^2]^2}\\ &=\frac{\frac1x(x^2)-2x(\ln x)}{x^4}\\ &=\frac{x-2x(\ln x)}{x^4}\\\\ &=\boxed{\frac{1-2\ln x}{x^3}} \end{align}$$