Please help, my live tutor and I could spent 2 hours trying to solve these two:
find an equation for tangent line y=-1-8^2@ (-4,-129)
find the derivative of the function:----we both kept getting the same answer, but Mathlab kept saying that it was wrong: -4lnl3xl/2+3x
we did the quotient rule, and the chain rule...the derivative and our sign would be wrong...
I cannot make sense out of your equation, check your typing.
-4 ln l3xl
2+3x
-4ln absolute value of 3x over 2+3x
y= -1-8x^2
(-4,-129)
the first set is in fraction form.
the 3x is absolute.
the 8x^2 is 8x squared.
for the derivative using the quotient rule I got (-8 - 12x + 12x(ln(3x))) / (x(2+3x)^2)
for the tangent question
dy/dx = -16x
at the point the slope is -16(-4) = 64
so using y = mx + b for the tangent equation
-129 = 64(-4) + b
b = 127
so the tangent equation is y = 64x + 127
Thanks. One small error? I can see what we did wrong..we did not do the
-129= part.....guess we were tired, or tried to do it in our heads.
To find the equation of the tangent line, we need to follow a few steps:
Step 1: Find the slope of the tangent line.
The slope of the tangent line at a particular point can be found by taking the derivative of the function at that point. In this case, we need to find the derivative of the function.
The function you provided is: y = -1 - 8^2.
However, the notation you used ("@") is not clear. It looks like you might have meant an exponent of 2. If that's the case, the function can be written as: y = -1 - 8^2 = -1 - 64 = -65.
Now, let's find the slope of the tangent line:
Step 2: Find the derivative of the function.
You mentioned that you were trying to find the derivative of the function -4ln|3x|/2+3x. Let's go through the process to find the derivative correctly.
To differentiate this function, we will use the quotient rule and the chain rule. The quotient rule states that if we have a function of the form f(x) = g(x)/h(x), then the derivative is given by: f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))^2.
In this case, let g(x) = -4ln|3x| and h(x) = 2 + 3x. We will need to find g'(x) and h'(x).
To find g'(x), we can apply the chain rule. The derivative of ln|3x| with respect to x is (1/3x) * 3 = 1/x.
Therefore, g'(x) = -4 * (1/x) = -4/x.
To find h'(x), we differentiate 2 + 3x, which is simply 3.
Now, we can substitute these values into the quotient rule formula to find the derivative of the function:
f'(x) = (-4/x)(2 + 3x) - (-4ln|3x|)(3) / (2 + 3x)^2
Simplifying further:
f'(x) = (-8 - 12x + 12ln|3x|) / (2 + 3x)^2
This is the correct derivative of the function.
Regarding the sign issue you mentioned, make sure you are applying the product rule and the chain rule correctly, and carefully track the signs of each term. Double-check your calculations to avoid any sign errors.
Now that we have the slope and the point (-4, -129), we can use the point-slope form of a line to find the equation of the tangent line.
The point-slope form is: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point.
Substituting the values: m = (-8 - 12(-4) + 12ln|3(-4)|) / (2 + 3(-4))^2 = -8 + 48 + 12ln12 / 14^2 = 40 + 12ln12 / 196.
Now, we substitute the values into the point-slope form:
y - (-129) = (40 + 12ln12 / 196)(x - (-4))
Simplifying:
y + 129 = (40 + 12ln12 / 196)(x + 4)
This is the equation of the tangent line.