Find a function f such that f'(x) = x^3 and the line x + y = 0 is tangent to the graph of f.

The problem states:

Find a function f such that f'(x) = x^3 and the line x + y = 0 is tangent to the graph of f.

Follow steps below to solve (C is the unknown constant):

Use x + y = 0 and solve for y we get y=-x:
Take the derivative of y and we see that y' = -1.
Since f'(x) = y', we can use y' = -1 to solve for x by saying -1 = x^3 and solving for x we get x = -1:
Using x = -1 in x + y = 0 we can find that y = 1:
Take the antiderivative  of f'(x) and we get f(x) = (1/4)x^4 + C
Because we know x and y, we can use those to solve for C:
Using C and the antiderivative gives us our final answer for our function f:

Find the derivative of y=10^(1-x^2)

Find the derivative of the function y = 10^(1-x^2).

Original equation:
To setup the chain rule, we will set the inner function as (1-x^2) and the outer function as (10^u) where u=1-x^2.
Using the chain rule, we take the derivative of the outer function times the derivative of the inner function:
Substituting 1-x^2 back in for u we get our final answer: