Find the equation of the tangent line to the curve y=x^4+2 e^x at the point (0,2)?

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Feb 10, 2018

Answer:

#y=2x+2#

Explanation:

Now, remember these rules:
#f(x)=g(x)+h(x)=>f'(x)=g'(x)+h'(x)#

#d/dx(e^x)=e^x#

#d/dx(nx)=n*d/dx(x)# where #n# is a constant.

#d/dx(x^n)=nx^(n-1)# where #n# is a constant.

It is important to keep in mind that #x# is the variable and #n# is a constant!

#y=x^4+2e^x#
=>#d/dx(y)=d/dx(x^4)+2*d/dx(e^x)#

=>#dy/dx=4x^3+2e^x#

We now plug in #0# in the place of #x#.

=>#dy/dx=4*0^3+2e^0#

=>#dy/dx=4*0^3+2e^0#

=>#dy/dx=0+2*1#

=>#dy/dx=2#

Now, we need to write the equation of the slope.

We use the following information: #y-y_1=m(x-x_1)#

We have: #y-2=2(x-0)#

=>#y=2x+2#

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mizoo Share
Feb 9, 2018

Answer:

#y = 2x + 2#

Explanation:

#y=x^4+2 e^x#
#y' = 4x^3 + 2e^x#

at #x = 0, y' = 4 xx 0 + 2e^0 = 2#

#m = 2, (0, 2)#
#y - 2 = 2(x - 0)#
#y = 2x + 2#

Here is the graph: https://www.wolframalpha.com/input/?i=plot+%7By%3Dx%5E4%2B2+e%5Ex,+y+%3D+2x+%2B+2%7D

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