Finding equation of tangent to the curve

Question

Assignment:

Q1. Find dy/dx for each of the following functions.

a) y=3x⁴-6x²+2x

dy/dx =3x³-6x+2

=9x-6

b) y= 3/x²

y=3x^-2

dy/dx =3x^-3

c) y=8x⁴-5x²-2/ 4x³

y=2x  -  5/4x   -1/2x³

y=2x - ???

d) y=square root 5x  -  square root x/5

e) y=(x²-6x+3)³

y=x⁶-216x³+27

dy/dx =x⁵-648x²

Q2. a) Find the slope of the tangent to the curve y=4x³-3x²+1 at the point where x=-1.

b) Find the equation of the tangent to the curve y=2x-x² that is perpendicular to the line x+4y-8=0.

Q3. If f(x)=320, find the derivative of f(x). Clearly give reason(s) for the answer.

Q4. Given f(x)=x³-1/2 x²+1. Find the value of x where dy/ dx =4.

Provide complete and step by step solution for the question and show calculations and use formulas.