Reflection 4

This week in calculus, we continued to learn about derivatives. Also, we learned how to be able to tell if a derivative is continuous and differentiable. To be able to tell if it is differentiable you must know your notes, and understand what a cusp, vertical tangent, etc. is. After we learned those things, we learned the shortcut for solving derivatives.

Examples:

Find the derivative. (* > delta)
f(x) = 2x^2
lim 2(x+*x)^2 - 2x^2 – 2x^2 + 4x*x + 2*x^2 -2x^2 / *x
*x>0
= *(4x+2*x) / *x
= 4x

Take the derivative of f(x) = absolute value(x-3) at (3, 0).
Set the inside equal to 0. x-3=0
+3+3
x=3
So there is no solution because it is not differentiable at x=3.

Find the derivatives and slope using the shortcuts.
1) y = x^3 (1, 1)
exponent goes in front of variable, and subtract exponent by 1.
y’(x) = 3x^3-1
= 3x^2
m = 3(1)^2 = 3

2) y = 1/x^5
y’(x) = 1/5x^5-1
= 1/5x^4