Logaritimos

(Videos) Derivatives of Logarithmic Functions

by PatrickJMT

Here's a video by patrickJMT where he goes through some examples of finding the derivative of a logarithmic function.

The rule for logarithmic differentiation are as follows: $$\begin{align} &\frac{d}{dx}\ln[x]=\frac1x\\ &\frac{d}{dx}\ln[f(x)]=\frac{1}{f(x)}\cdot f'(x)\\ &\frac{d}{dx}\log_a(x)=\frac{1}{x\cdot\ln a}\\ &\frac{d}{dx}\log_a[f(x)]=\frac{1}{f(x)\cdot\ln a}\cdot f'(x) \end{align}$$

Remember to also use the properties of logarithms when taking derivatives. They can help turn a complicated problem into a less complicated problem.

The first problem he looks at is: $$f(x)=\ln(x^2+10)$$ So we take one over the inside and multiply that by the derivative of the inside. $$\begin{align} f'(x)&=\frac{1}{x^2+10}\cdot\color{red}{(x^2+10)'}\\ &=\frac{1}{x^2+10}\cdot2x\\ &=\boxed{\frac{2x}{x^2+10}} \end{align}$$

The second problem is $$f(x)=\sqrt x\cdot\ln x$$ Here we need to use the product rule to find this derivative. $$\begin{align} f'(x)&=\color{red}{(\sqrt x)'}\ln x+\sqrt x\color{red}{(\ln x)'}\\ &=\frac12x^{-\frac12}\ln x+\sqrt x\bigg(\frac1x\bigg)\\ &=\frac{\ln x}{2\sqrt{x}}+\frac{\sqrt x}{x}\\ &=\frac{\ln x}{2\sqrt{x}}+\frac{1}{\sqrt x}\\ &=\frac{\ln x}{2\sqrt{x}}+\frac{2}{2\sqrt x}\\ &=\boxed{\frac{2+\ln x}{2\sqrt{x}}} \end{align}$$

The third problem is: $$f(x)=\ln\bigg[\frac{(2x+1)^3}{(3x-1)^4}\bigg]$$ This is an example of a problem where it would be made easier if we used a logarithmic property. So let's use: $$\ln\bigg(\frac ab\bigg)=\ln a-\ln b\\ \ln a^n=n\ln a$$ $$\begin{align} f(x)&=\ln(2x+1)^3-\ln(3x-1)^4\\ &=3\ln(2x+1)-4\ln(3x-1)\\ f'(x)&=3\bigg(\frac{1}{2x+1}\bigg)\color{red}{(2x+1)'}-4\bigg(\frac{1}{3x-1}\bigg)\color{red}{(3x-1)'}\\ &=\boxed{\frac{6}{2x+1}-\frac{12}{3x-1}} \end{align}$$

The final example is: $$y=[\log_4(1+e^x)]^2$$ Here we will need to use the chain rule first. $$\begin{align} y'&=2\log_4(1+e^x)\cdot\color{red}{[\log_4(1+e^x)]'}\\ &=2\log_4(1+e^x)\cdot\frac{1}{(1+e^x)\cdot\ln 4}\cdot\color{red}{(1+e^x)'}\\ &=2\log_4(1+e^x)\cdot\frac{1}{(1+e^x)\cdot\ln 4}\cdot e^x\\ &=\boxed{\frac{2e^x\log_4(1+e^x)}{(1+e^x)\cdot\ln 4}} \end{align}$$

Here's another video with some more examples.

The first example he looks at: $$\begin{align} y&=\ln(x^2+x)\\ y'&=\frac{1}{x^2+x}\cdot\color{red}{(x^2+x)'}\\ &=\frac{1}{x^2+x}\cdot(2x+1)\\ &=\boxed{\frac{2x+1}{x^2+x}} \end{align}$$

The next problem is: $$y=\sqrt[3]{\log_7 x}$$ Here we have a log inside a chain rule. But first let's rewrite the problem without the radical symbol. $$\begin{align} y&=(\log_7 x)^{\frac13}\\ y'&=\frac13(\log_7 x)^{-\frac23}\cdot\color{red}{(\log_7 x)'}\\ &=\frac13(\log_7 x)^{-\frac23}\cdot\frac{1}{x\cdot\ln7}\\ &=\boxed{\frac{1}{3(\log_7 x)^{\frac23}\cdot x\cdot\ln7}} \end{align}$$

The third problem is: $$y=\ln(x^4\cdot\sin x)$$ Here we have a chain rule inside a logarithm...or we can use a logarithm property to make everyone's lives easier. $$\begin{align} y&=\ln(x^4)+\ln(\sin x)\\ &=4\ln x+\ln(\sin x)\\ y'&=4\bigg(\frac1x\bigg)+\frac{1}{\sin x}\cdot\color{red}{(\sin x)'}\\ &=\frac4x+\frac1{\sin x}\cdot(\cos x)\\ &=\frac4x+\frac{\cos x}{\sin x}\\ &=\boxed{\frac4x+\cot x} \end{align}$$

The final example is: $$y=\frac{\ln x}{1+\ln(2x)}$$ Here we have logarithms inside a quotient rule. $$\begin{align} y'&=\frac{\color{red}{(\ln x)'}[1+\ln(2x)]-\ln x\color{red}{[1+\ln(2x)]'}}{[1+\ln(2x)]^2}\\ &=\frac{\frac{1+\ln(2x)}{x}-\ln x[\frac{1}{2x}\cdot\color{red}{(2x)'}]}{[1+\ln(2x)]^2}\\ &=\frac{\frac{1+\ln(2x)}{x}-\ln x[\frac{2}{2x}]}{[1+\ln(2x)]^2}\\ &=\frac{\frac{1+\ln(2x)}{x}-\frac{\ln x}{x}}{[1+\ln(2x)]^2}\\ &=\frac{\frac{1+\ln(2x)-\ln x}{x}}{[1+\ln(2x)]^2}\\ &=\frac{1+\ln(2x)-\ln x}{x[1+\ln(2x)]^2}\\ &=\frac{1+\ln(\frac{2x}{x})}{x[1+\ln(2x)]^2}\\ &=\boxed{\frac{1+\ln2}{x[1+\ln(2x)]^2}}\\ \end{align}$$

(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

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

(Videos) Derivatives of Logarithms

by PatrickJMT

Here's a video by patrickJMT where he shows you how to take the derivative of logarithmic functions.

The two derivative rules you need to know are for natural logs and regular logs. $$\frac{d}{dx}\ln[f(x)]=\frac{1}{f(x)}\cdot f'(x)\\ \frac{d}{dx}\log_a[f(x)]=\frac{1}{f(x)\cdot\ln a}\cdot f'(x)$$

The first derivative problem he looks at is: $$y=\ln x+2^x+\sin x$$ This is a very straight forward problem that only requires you to know the different derivative rules. $$\begin{align} y'&=\boxed{\frac1x+2^x\cdot\ln2+\cos x} \end{align}$$

The next problem is: $$h(x)=\ln(x^2+3x+4)$$ Here we need to use the chain rule when taking the derivative of the natural log. $$\begin{align} h'(x)&=\frac{1}{x^2+3x+4}\cdot\color{red}{(x^2+3x+4)'}\\ &=\frac{1}{x^2+3x+4}\cdot(2x+3)\\ &=\boxed{\frac{2x+3}{x^2+3x+4}} \end{align}$$

This next video is a continuation of the first. It has two more logarithmic derivative problems that he solves.

The first problem of this video is: $$g(x)=\log_4(x^3+8x)$$ Here we have a chain rule paired with a logarithm with base four. $$\begin{align} g'(x)&=\frac{1}{(x^3+8x)\cdot\ln4}\cdot\color{red}{(x^3+8x)'}\\ &=\frac{1}{(x^3+8x)\cdot\ln4}\cdot(3x^2+8)\\ &=\boxed{\frac{3x^2+8}{(x^3+8x)\cdot\ln4}} \end{align}$$