Find an equation of the tangent line to the given curve at thespecified point y=exx;

Oberlaudacu

Oberlaudacu

Answered

2021-12-23

Find an equation of the tangent line to the given curve at thespecified point
y=exx; (1,ex)

Answer & Explanation

Jenny Sheppard

Jenny Sheppard

Expert

2021-12-24Added 35 answers

Consider the curve y=exx
The objective is to find an equation of the tangent line to the given curve at the point (1,ex)
Consider the curve y=f(x). Then, equation of tangent at the point (x1, y1) is given by, yy1=m(xx1)
Here, m=dydx(x1, y1) =slope of the tangent at the point (x1, y1).
In the given question. f(x)=exx and (x1, y1)=(1, ex).
Differentiate y=exx with respect to x:
dydx=ddx(exx)
=xddx(ex)exddx(x)x2
=xexex(1)x2
=ex(x1)x2
Evaluate dydx at (x1, y1)=(1, ex):
m=dy1dx1=e1(11)12=0
Thus, equation of tangent line is:
yy1=m(xx1)
yex=0(x1)
ye1=0 (at x=1)
y=e
Therefore, equation of the tangent line to the given curve at the point (1, ex) is y=e

Maricela Alarcon

Maricela Alarcon

Expert

2021-12-25Added 28 answers

Possible derivation:
ddx(exx)
Use the quotient rule, ddx(uv)=vdudxudvdxv2, where u=ex and v=x:
=x(ddx(ex))ex(ddx(x))x2
Using the chain rule, ddx(ex)=deudududx, where u=x and ddu(eu)=eu:
=ex(ddx(x))+ex(ddx(x))xx2
The derivative of x is 1:
=ex(ddx(x))+1exxx2
The derivative of x is 1:
=exx1exx2
Simplify the expression:
Answer: =ex(1+x)x2
user_27qwe

user_27qwe

Expert

2021-12-30Added 230 answers

First answer is right!

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