{
"query": {
"display": "derivative of $$y=\\ln\\left(2x\\right)$$",
"symbolab_question": "PRE_CALC#derivative y=\\ln(2x)"
},
"solution": {
"level": "PERFORMED",
"subject": "Calculus",
"topic": "Derivatives",
"subTopic": "Derivatives",
"default": "\\frac{1}{x}",
"meta": {
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}
},
"steps": {
"type": "interim",
"title": "$$\\frac{d}{dx}\\left(\\ln\\left(2x\\right)\\right)=\\frac{1}{x}$$",
"input": "\\frac{d}{dx}\\left(\\ln\\left(2x\\right)\\right)",
"steps": [
{
"type": "interim",
"title": "Apply the chain rule:$${\\quad}\\frac{1}{2x}\\frac{d}{dx}\\left(2x\\right)$$",
"input": "\\frac{d}{dx}\\left(\\ln\\left(2x\\right)\\right)",
"result": "=\\frac{1}{2x}\\frac{d}{dx}\\left(2x\\right)",
"steps": [
{
"type": "step",
"primary": "Apply the chain rule: $$\\frac{df\\left(u\\right)}{dx}=\\frac{df}{du}\\cdot\\frac{du}{dx}$$",
"secondary": [
"$$f=\\ln\\left(u\\right),\\:\\:u=2x$$"
],
"result": "=\\frac{d}{du}\\left(\\ln\\left(u\\right)\\right)\\frac{d}{dx}\\left(2x\\right)",
"meta": {
"practiceLink": "/practice/derivatives-practice#area=main&subtopic=Chain%20Rule",
"practiceTopic": "Chain Rule"
}
},
{
"type": "interim",
"title": "$$\\frac{d}{du}\\left(\\ln\\left(u\\right)\\right)=\\frac{1}{u}$$",
"input": "\\frac{d}{du}\\left(\\ln\\left(u\\right)\\right)",
"steps": [
{
"type": "step",
"primary": "Apply the common derivative: $$\\frac{d}{du}\\left(\\ln\\left(u\\right)\\right)=\\frac{1}{u}$$",
"result": "=\\frac{1}{u}"
}
],
"meta": {
"solvingClass": "Derivatives",
"interimType": "Derivatives",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s79Kg+idP5vLVrjUll6eMdYoqTCAmruKWcJsn66ZPDMT8cjlLRK1jUV206qo4+vRN78rEus7TgCihQBF5omOFkJq1PlbV5jLoKv9solFCc4blTW26qciuyUBGXQExCUedYd9mDo5FIvzrirtH7/W8pPUxk6YPA4jUd3Af4X0JJJ64="
}
},
{
"type": "step",
"result": "=\\frac{1}{u}\\frac{d}{dx}\\left(2x\\right)"
},
{
"type": "step",
"primary": "Substitute back $$u=2x$$",
"result": "=\\frac{1}{2x}\\frac{d}{dx}\\left(2x\\right)"
}
],
"meta": {
"interimType": "Derivative Chain Rule 0Eq",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s79Kg+idP5vLVrjUll6eMdYjgc+L3QSQmb0rhR14S/trj8zeERICEnv1Ds5A1/BdIw2AW2hhsmfO/7M0RMNsJBcxhgKq+X573ilbTfYIBcNAzWM9HA0a5VHxu5rWRbGwYViSshf24Xf2esyeeKCdsFxiIim3Xa07T8DZYj24lqtW4EuDOVaQvKofqHoY5jNapszx1SRmkdwriKeTV96yww0ImpXFf3SOUx+H18qfp3MLg="
}
},
{
"type": "interim",
"title": "$$\\frac{d}{dx}\\left(2x\\right)=2$$",
"input": "\\frac{d}{dx}\\left(2x\\right)",
"steps": [
{
"type": "step",
"primary": "Take the constant out: $$\\left(a{\\cdot}f\\right)'=a{\\cdot}f'$$",
"result": "=2\\frac{dx}{dx}"
},
{
"type": "step",
"primary": "Apply the common derivative: $$\\frac{dx}{dx}=1$$",
"result": "=2\\cdot\\:1"
},
{
"type": "step",
"primary": "Simplify",
"result": "=2",
"meta": {
"solvingClass": "Solver"
}
}
],
"meta": {
"solvingClass": "Derivatives",
"interimType": "Derivatives",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s79Kg+idP5vLVrjUll6eMdYg2sQzwGEAAPyDk8n13Ps8XZGku9zFkxwe1dTH8vycb94wHsFp27x8BxzSfXYcuPllNbbqpyK7JQEZdATEJR51jH4j/fzMjnIhJwos1vPNWw"
}
},
{
"type": "step",
"result": "=\\frac{1}{2x}\\cdot\\:2"
},
{
"type": "interim",
"title": "Simplify $$\\frac{1}{2x}\\cdot\\:2:{\\quad}\\frac{1}{x}$$",
"input": "\\frac{1}{2x}\\cdot\\:2",
"result": "=\\frac{1}{x}",
"steps": [
{
"type": "step",
"primary": "Multiply fractions: $$a\\cdot\\frac{b}{c}=\\frac{a\\:\\cdot\\:b}{c}$$",
"result": "=\\frac{1\\cdot\\:2}{2x}"
},
{
"type": "step",
"primary": "Cancel the common factor: $$2$$",
"result": "=\\frac{1}{x}"
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Generic Simplify Specific 1Eq",
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}
}
],
"meta": {
"solvingClass": "Derivatives",
"practiceLink": "/practice/derivatives-practice",
"practiceTopic": "Derivatives"
}
},
"plot_output": {
"meta": {
"plotInfo": {
"variable": "x",
"funcsToDraw": {
"funcs": [
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"evalFormula": "y=\\frac{1}{x}",
"displayFormula": "y=\\frac{1}{x}",
"derivativeFormula": "-\\frac{1}{x^{2}}",
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"labels": [
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Solution
derivative of
Solution
Solution steps
Apply the chain rule:
Simplify
Graph
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Frequently Asked Questions (FAQ)
What is the derivative of y=ln(2x) ?
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