Find an equation of the tangent line to the given curve at thespecified point
$y = \frac{e^{x}}{x}; (1, e^{x})$

Question

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

Jenny Sheppard · Accepted Answer

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, y−y1=m(x−x1) 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 =xex−ex(1)x2 =ex(x−1)x2 Evaluate dydx at (x1, y1)=(1, ex): m=dy1dx1=e1(1−1)12=0 Thus, equation of tangent line is: y−y1=m(x−x1) y−ex=0(x−1) y−e1=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 · Accepted Answer

Possible derivation:  ddx(exx)  Use the quotient rule, ddx(uv)=vdudx−udvdxv2, 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:  =exx−1exx2  Simplify the expression:  Answer: =ex(−1+x)x2

user_27qwe · Accepted Answer

First answer is right!

Find an equation of the tangent line to the given

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