Exponential Derivatives (1 Viewer)

[taggs-sasuke]

Find the equation of the tangent to the curve y = e^x which passes through the origin. Ans: y = ex.

I can't get the answer.

 

[shaon0]

Find the equation of the tangent to the curve y = e^x which passes through the origin. Ans: y = ex.

I can't get the answer.

y=e^x-1
y'=e^x

 

[taggs-sasuke]

y=e^x-1

I'm sorry. How did you get this?

 

[xMrRand0m]

I'm sorry. How did you get this?

The original curve that you gave, y=e^x does not pass through the origin, the curve y=e^x-1 does.

 

[Dumsum]

Forget y=e^x-1, they've misunderstood the question.

y = e^x
y' = e^x

So at point (k,e^k), tangent has gradient e^k.

Use point-gradient formula:
y - e^k = e^k(x - k)

For different values of k, this gives the equation of different tangents... We want the one that passes through the origin so we need to solve the above equation for k at (x,y) = (0,0).

-e^k = -ke^k
e^k (k - 1) = 0
k = 1 (since e^k can never be 0).

Substituting this k value gives
y - e = e(x - 1)
y = e + ex - e
y = ex.

 

[xMrRand0m]

Forget y=e^x-1, they've misunderstood the question.

y = e^x
y' = e^x

So at point (k,e^k), tangent has gradient e^k.

Use point-gradient formula:
y - e^k = e^k(x - k)

For different values of k, this gives the equation of different tangents... We want the one that passes through the origin so we need to solve the above equation for k at (x,y) = (0,0).

-e^k = -ke^k
e^k (k - 1) = 0
k = 1 (since e^k can never be 0).

Substituting this k value gives
y - e = e(x - 1)
y = e + ex - e
y = ex.

Ooh! My mistake then, I thought she meant find the tangent of y=e^x-1 at the origin. Sorry.

 

[taggs-sasuke]

Forget y=e^x-1, they've misunderstood the question.

y = e^x
y' = e^x

So at point (k,e^k), tangent has gradient e^k.

Use point-gradient formula:
y - e^k = e^k(x - k)

For different values of k, this gives the equation of different tangents... We want the one that passes through the origin so we need to solve the above equation for k at (x,y) = (0,0).

-e^k = -ke^k
e^k (k - 1) = 0
k = 1 (since e^k can never be 0).

Substituting this k value gives
y - e = e(x - 1)
y = e + ex - e
y = ex.

Oh, I get it. Thank you!

 

[shaon0]

Ooh! My mistake then, I thought she meant find the tangent of y=e^x-1 at the origin. Sorry.

Me to.