6) Find derivative of y=2

7)Find equation of tangent line to the graph of x^2+2y^2=3 at point of (1,1)

8)let g(x)=9f(x) and left f'(-6)=-6 find g'(-6)

9) Find derivative of x^-9

10) Find the value of the derivative f(t)=t^3+2/t at point (-2,3)

6) Answer: d(y)/dx = d(2)/dx = 0

7) x^2 + 2y^2 = 3

2x + 4y * dy/dx = 0

4y * dy/dx = -2x

dy/dx = -2x/4y = -x/2y

At the point (1, 1)

dy/dx = (-1)/(2 * 1) = -1/2

The equation of the tangent line is:

y - 1 = (-1/2)(x - 1)

y - 1 = -x/2 + 1/2

Answer: y = -x/2 + 3/2

8) g(x) = 9*f(x), f'(-6) = -6, g'(-6) = ?

g'(x) = 9 * f'(x)

g'(-6) = 9 * f'(-6)

Answer: g'(-6) = 9 * -6 = -54

9) y = x^-9

Answer: dy/dx = -9x^-10

10) f(t) = t^3 + 2/t = t^3 + 2t^-1

f'(t) = 3t^2 - 2t^-2 = 3t^2 -2/t^2

At the point (-2, 3)

f'(-2) = 3*(-2)^2 - 2/(-2)^2

Answer: f'(-2) = 12 - 1/2 = 23/2

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up