You use the quotient rule for derivatives since you are dividing.
We will use F(x) for sinx and G(x) for (1+tanx), and H'(x) for the final answer.
So H'(x) = (F'(x)G(x) - F(x)G'(x))/(G(x))^(2))
The derivative of sinx is cosx so F'(x) = cosx
The derivative of 1+tanx is sec^(2)x, just like the derivative of tangent (because the derivative of a constant is 0, and by the addition rule for derivatives you would add the derivatives). 0 + sec^(2)x is sec^(2)x. So G'(x) is sec^(2)x.
So with this information just plug F, F', G, and G' into the H'(x) = (F'(x)G(x) - F(x)G'(x))/(G(x))^(2)) equation and solve.
H'(x) = (cosx)(1+tanx) - (sinx)(sec^(2)x)/((1+tanx)^(2))
Answer:
A
Step-by-step explanation:
Answer:
y = x + 3
Step-by-step explanation:
I'm assuming that you are asking for a line perpendicular to
3x + 3y = 3 and going through (-1, 2)
first find the slope of the line
3y = -3x + 3
y = -1x + 1, the slope is -1 so the perpendicular slope is 1
y = mx + b
2 = 1(-1) + b
3 = b
y = x + 3
4/5x=7/5 you have to divide 7/5 by 4/5 which equals to 1.75/5 which is equal to x
