JEE Exam > JEE Questions > The equation of the tangent to the curvey = e...
Start Learning for Free
The equation of the tangent to the curve y = e2x at the point (0, 1) is
- a)1 – y = 2 x
- b)y – 1 = 2 x
- c)y + 1 = 2 x
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The equation of the tangent to the curvey = e2x at the point (0, 1) is...
Hence equation of tangent to the given curve at (0 , 1) is :
(y−1) = 2(x−0),i.e..y−1 = 2x
Most Upvoted Answer
The equation of the tangent to the curvey = e2x at the point (0, 1) is...
Finding the Derivative of y=e^(2x)
To find the equation of the tangent to the curve y = e^(2x) at the point (0,1), we first need to find the derivative of the function y = e^(2x) using the power rule of differentiation.
dy/dx = 2e^(2x)
Finding the Equation of the Tangent Line
Now that we have the derivative of the function, we can find the equation of the tangent line to the curve at the point (0,1) using the point-slope form of a line.
y - y1 = m(x - x1)
where m is the slope of the tangent line and (x1, y1) is the point on the curve where we want to find the tangent line.
Substituting x1 = 0 and y1 = 1, and m = dy/dx evaluated at (0,1), we get
y - 1 = 2e^(2*0)(x - 0)
y - 1 = 2x
Simplifying the equation, we get
y = 2x + 1
Therefore, the equation of the tangent to the curve y = e^(2x) at the point (0,1) is y - 1 = 2x. Option 'B' is the correct answer.
To find the equation of the tangent to the curve y = e^(2x) at the point (0,1), we first need to find the derivative of the function y = e^(2x) using the power rule of differentiation.
dy/dx = 2e^(2x)
Finding the Equation of the Tangent Line
Now that we have the derivative of the function, we can find the equation of the tangent line to the curve at the point (0,1) using the point-slope form of a line.
y - y1 = m(x - x1)
where m is the slope of the tangent line and (x1, y1) is the point on the curve where we want to find the tangent line.
Substituting x1 = 0 and y1 = 1, and m = dy/dx evaluated at (0,1), we get
y - 1 = 2e^(2*0)(x - 0)
y - 1 = 2x
Simplifying the equation, we get
y = 2x + 1
Therefore, the equation of the tangent to the curve y = e^(2x) at the point (0,1) is y - 1 = 2x. Option 'B' is the correct answer.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer?
Question Description
The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer?.
The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer?.
Solutions for The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The equation of the tangent to the curvey = e2x at the point (0, 1) isa)1 – y = 2 xb)y – 1 = 2 xc)y + 1 = 2 xd)none of theseCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup