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 =

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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 =

Mathematics
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can you find the derivative of the curve...?
i don't know
ok... have you done or are you doing calculus?

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Other answers:

doing calculus
ok... can you differentiate y = x^4...?
4x^3
and can you differentiate \[ y = 4e^{x} \]
x4e
or is it x4e^-x
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 \[y = 4e^x\]
4xe^-x
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)
y'=4
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..?
=-1/4x^3+4e^x
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...?
y=4x-4
can you try again..
y=4x+4
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..?
y=-1/4x+4
that looks good... so there you have your solutions.
thank you
hope it helps... you do need to practice you differentiation basics.
yeaa

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