I have the answers to this study guide but if someone can work out the problems

ID: 2848361 • Letter: I

Question

I have the answers to this study guide but if someone can work out the problems and show me how it's done, that would be great!!

1. If R(x) = (x^3 + 3x)^5 find R'(x)

(This answer is: 5(3x^2+3)(x^3+3x)^4)

2. If z=5t^3 + e^-1.5t, find dz/dt

(This answer is: 15t^2 - 1.3e^-1.3t)

3. If q=ln(s^3+s) find dq/ds

(This answer is: dq/ds = 3s^2+1/s^3+s)

4. If y=x^2ln(x), find dy/dx

(This answer is: 2xln(x) + x)

5. Find the equation of the line tangent to the graph of f(x)=2x^3-5x^2+3x-5 at x=1

(This answer is: y=-x-4)

Thanks in advance for showing me how to do it!!!!

you need to know the formula d/dx(x^n) = n*x^(n-1)

R(x) = (x^3 + 3x)^5

R'(x)= 5*(x^3 + 3x)^(5-1) * d/dx(x^3 + 3x)

R'(x) = 5*(x^3+3x)^4 *(3x^2 + 3)

z=5t^3 + e^-1.5t

d/dx(e^x) = e^x

dz/dt = 3*5*t^(3-1) + e^(-1.5t) * d/dt(-1.5t)

dz/dt = 15*t^2 + e^(-1.5t) *(-1.5)

dz/dt = 15t^2 - 1.5*e^(-1.5t)

q=ln(s^3+s)

d/dx(lnx) = 1/x

dq/ds = 1/(s^3+s) * d/ds(s^3+s)

dq/ds = 3s^2 + 1/(s^3+s)

y=x^2ln(x)

d(uv) = udv + vdu

dy/dx = x^2/x + lnx * 2x

dy/dx = x + 2xlnx

f(x)=2x^3-5x^2+3x-5 at x=1

df/dx = 6x^2 -10x + 3

at x = 1, m = 6 - 10 + 3 = -1

y = mx + c

2-5+3-5 = -1 + c

c = -4

so, eqn is y = -x-4