Basic Rules
What is the product rule?
d/dx (f(x) * g(x)) = f'(x)g(x) + f(x)g'(x)
d/dx(cosx)=
-sinx
d/dx(ln(x))
1/x
What rule would you use to find the derivative of cos^4(x)?
A) chain rule, B) product rule, C) trig product rule, D) L'hopital's rule
A) chain rule
What is the quotient rule for derivatives?
d/dx[f(x)/g(x)] = [g(x)f'(x) - f(x)g'(x)] / (g(x))^2
OR
[ho*d(hi) - hi*d(ho)] / (ho*ho
d/dx(-sin^2x)=
-2sinxcosx
d/dx(b^x)
b^x*ln(b)
Given x^3-y^2=0, evaluate dy/dx at the point (1,-1)
dy/dx|(1,-1) = -3/2
What is the chain rule for derivatives?
d/dx (f(g(x)) = f'(g(x))*g'(x)
d/dx(cos^3(4x))
-12cos^2(4x)sin(4x)
d/dx(2xe^(2x))=
4xe^(2x)+2e^(2x)
If f(x)=(x^2-2x-1)^(2/3), then f'(0)=
4/3
f(x)=(4x^2-3x+2)^-2
f'(x)=
f'(x)=(-2(8x-3))/(4x^2-3x+2)^3
If f(x)=x/tanx, find f'(π/4)
1-pi/2
If f(x)=3^cosx, find f'(π/2)
f'(pi/2)=-ln3
-7/(sqrt(1-49x^2))+(6lnx)/x
-7/(sqrt(1-49x^2))+(6lnx)/x
f(x)=(3sinx+lnx)/(x^3+2x)
f'(x)=
f'(x)=((x^3+2x)(3cosx+1/x)-(3sinx+lnx)(3x^2+2))/((x^3+2x)^2
d/dx(-secxcotx)=
cscxcotx
Only need it in one term, not completely simplfied
d/dx(e^(3ln(4x^2+1)))
(24xe^3ln(4x^2+1))/(4x^2+1)
d/dx((-12x^4-16x^3+12x^2)/(4x^6))
(96x^9+192x^8-192x^7)/(4x^12)