Find an equation of the tangent line to the curve at the given point.
y= (1+3x)^12 , (0,1)
and find y' and y"
y=cos(x^2)
I got y' = -2xsin(x^2)
i keep trying to get double prime but everything i try isnt working.
first:
dy/dx = 12(1+3x)^11 (3)
= 36(1+3x)^11
when x = 0
dy/dx = 36(1)^11 = 36
and of course (0,1) is the y-intercept, so equation is
y = 36x + 1
2nd:
y = cos(x^2)
y' = 2x(-sin(x^2) = -2x sin(x^2)
use the product rule for y''
y'' = (-2x)cos(x^2) (2x) + sin(x^2) ( -2)
= -4x^2 cos(x^2) - 2sin(x^2)
I've tried entering in 36x+1 earlier and it told me it was incorrect. I thought I did it wrong. Thanks for getting back to me.
well i entered it in again and now it took that first answer. i think theres a glitch with the program but thank you for showing me how you got the answers. it helps so much!
To find an equation of the tangent line to the curve at the given point, you can follow these steps:
1. Differentiate the given equation with respect to x to find the derivative, which represents the slope of the tangent line at any point on the curve.
Let's start with the first problem:
y = (1 + 3x)^12
Differentiating both sides with respect to x, we use the chain rule:
dy/dx = 12(1 + 3x)^11 * 3
Simplifying further:
dy/dx = 36(1 + 3x)^11
Since we are evaluating the tangent line at the point (0,1), substitute x=0 into dy/dx:
dy/dx = 36(1 + 3(0))^11
dy/dx = 36(1)^11
dy/dx = 36
So, the slope of the tangent line is 36.
2. Use the point-slope form of a line to find the equation of the tangent line.
Using the point-slope form, we have:
y - y1 = m(x - x1)
Substituting the values (0,1) for (x1,y1) and 36 for m:
y - 1 = 36(x - 0)
y - 1 = 36x
Simplifying the equation, we get:
y = 36x + 1
Therefore, the equation of the tangent line is y = 36x + 1.
Now moving on to the second problem:
y = cos(x^2)
To find the first derivative:
Using the chain rule, we differentiate both sides with respect to x:
dy/dx = -sin(x^2) * 2x
Simplifying further:
dy/dx = -2xsin(x^2)
So, you were correct in finding the first derivative as -2xsin(x^2).
To find the second derivative, we differentiate the first derivative with respect to x again:
Using the product rule this time, we have:
d/dx(-2xsin(x^2)) = -2sin(x^2) - 4x^2cos(x^2)
Therefore, the second derivative (y") of y = cos(x^2) is:
y" = -2sin(x^2) - 4x^2cos(x^2)
Hope that helps!