find the equation of the tangent and the equation of the normal to the curve y = x - 1/3 x^3 + 1/7 x^5 where x = 4

The tangent line to the given curve, at the point x = 4 is
the derivative of the function at x = 4.

Therefore, we need
to determine the equation of the 1st derivative of the
function.

If the function is y = x - 1/(3x^3) + 1/(7x^5),
we'll get:

y' = 1 - (-9x^2)/9x^6 +
(-35x^4)/49x^10

y' = 1 + 1/x^4 -
5/7x^6

We'll calculate the slope of the tangent line, at
the point x = 4:

y'(4) = 1 + 1/256 -
5/7*4096

y'(4) = (28672 +112 - 5
)

y'(4) = 28779/28672

We'll
calculate the value of the function at x = 4:

y(4) = 4 -
1/192 + 1/7168

y(4) = (4*7168 - 112 +
3)/21504

y(4) =
28563/21504

Therefore, the equation of the
tangent line at the curve y = x - 1/(3x^3) + 1/(7x^5) t x = 4, is y - 28563/21504 =
28779(x-4)/28672.

The equation
of the normal line is written: y - 28563/21504 =
-28672(x-4)/28779.