Question

Find equations of the tangent line and normal line to the curve at the given point.
y = x^4 + 4e^x, (0, 4)
tangent line y =
normal line y =

Answer 1

can you find the derivative of the curve...?

Answer 2

i don't know

Answer 3

ok... have you done or are you doing calculus?

Answer 4

doing calculus

Answer 5

ok... can you differentiate y = x^4...?

Answer 6

4x^3

Answer 7

and can you differentiate

\[ y = 4e^{x} \]

Answer 8

x4e

Answer 9

or is it x4e^-x

Answer 10

quick lesson

\[y = 4e^{f(x)} ..... then .... y'=4f'(x) e^{f(x)}\]

and in your question f(x) = x...

try the derivative again for

\[y = 4e^x\]

Answer 11

4xe^-x

Answer 12

not quite

if f(x) = x

f'(x) = 1

so its

y' = 4e^x

so you you function the derivative is

\[y' = 4x^3 + 4e^x\]

ok... you now have the equation for the slop of the tangent...(1st derivative)

subsitiute x = 0 to find the slope of the tangent at the point (0, 4)

Answer 13

y'=4

Answer 14

thats correct so you have a tangent or line with a slope of 4 and point (0, 4)

can you find the equation of the line...?

for the normal... the slope will be the negative reciprocal of the slope of the tangent... what would the slope of the normal be..?

Answer 15

=-1/4x^3+4e^x

Answer 16

no... ok... lets step back...

you have a slope of 4 and point (0, 4)

the equation of the tangent will be

y = 4x + b

or using the formula

\[y - 4 = 4(x - 0)\]

whats the equation of the tangent...?

Answer 17

y=4x-4

Answer 18

can you try again..

Answer 19

y=4x+4

Answer 20

ok... thats the equation of the tangent...

now the normal... which is perpendicular to the tangent...

the slope is the negative reciprocal of the tangent...

so if the tangent has a slope of 4 ... what is the slope of the normal..?

Answer 21

y=-1/4x+4

Answer 22

that looks good... so there you have your solutions.

Answer 23

thank you

Answer 24

hope it helps... you do need to practice you differentiation basics.

Answer 25

yeaa