Description
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)
Explanation & Answer
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
Review
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