5 Calc I problems1. Demonstrate how the use the rules of derivatives to differentiate.f(x)= sin x cos^3x 2. same as aboveQ(x)= sinx/x^2+8 3. Find...

We use the following rules/ frmulas tind the derivatives:

d/dx (kx^n) = (kx^n)'= knx^(n-1)

d/dx(UV) = UV'+U'V ,

d/dx{f(u(x))} = f'(u(x))*du(x)/dx = f'(u(x))u'(x)

d/dx constant or number = d/dx k = d/dx*kx^0 = 0 and

d/dx{f(x)   +or-  g(x)}  = f'(x)  +or-  g'(x).

1) d/dx  f(x)= sin x cos^3x

=sinx(cos^3x)'+(sinx)' (cos3^x)

=sinx(cosx)^(3-1)*(-sinx)+cosx*cos^3 x

=-(sinx)^2 (cosx)^2+(cosx)^4

2)

d/dx Q(x)= sinx/x^2+8

=[(sinx)(x^(-2))]'+[8*x^0]'

=sinx(-2*x^(-2-1))+(cosx)*x^(-2) + 8*0

=-2x^(-3)*sinx+x^(-2)cosx

=-(2sinx)/x^3+(cosx)/x^2

3)Tind equation of f(x)= 2(-2x+6)^2   when x=5

The tangent equation is:

y = y'(x-x1) + y1, where  X1 =5 given and y1 = value of y at x=5.

y' at x=5  is 2(-2x+6)^(2-1)*(-2) at x=5

=2(-2x+6)(-2) at x=5

= 8x-24 at x=5

=40-24 =16 is the value of y' at x=5.

y1 = y at x=5

=2(-2x+6)^2 at x=5

=2(-10+6)^2

=32 is the value of y1 or y at x=5.

Therefore equation of the tangent: y = 16(x-5)+32  Or y= 16x-48 is the tangent.

4. FInd the slope of the tangent line to the graph of the function at the given value.
f(x)= -2/x^4  when x=7.

The slope of the tangent is y' =( -2/x^4)' at x=7

=[-2*x^(-4)]' at x=7

=(-2)(-4)x^(-4-1) at x=7

=8(7)^(-5). Very small slope and is the tagent of an angle, 0.027272339 degree with X axis.

5. Determine the point(s), (if any), at which the graph of the function has a horizontal tangent.
y(x)= x^3+15x^2+6

Solution:

A horizontal tangent has the slope zero.

y'(x) = (x^3+15x^2+6)' = (x^3)'+(15x^2)'+(6)'

=3x^2+15*2x

= 3x(x^2+5) is equal to zero when 3x = 0 Or when x = 0

When x=0 , y=6.

Therefore the tangent y =y'(x-x1)+ y1 or the value of y at x=0 becomes:

y=0*(x-5)+ 6 Or

y = 6 is the horizontal tangent.