Derivative questions?
1. Compute the derivative. d/dx [(x^3+9x)(x^2-x)] |x=2 I solved the derivative first: (3x+9)(x^2-x)+(x^3+9x)(2x-1) = 2x^4 - 2x^3 + 24x^2 - 18x. Then I plugged in 2 and got 76 and it's wrong. What am I doing wrong? 2. Find the indicated derivative. In this case, the independent variable is a (unspecified) function of t. s = 8r + r^-1. Find ds/dt. ds/dt = (__?__) dr/dt I'm confused on what it's asking me. How would I find t? 3. Compute the indicated derivative using the chain rule. y = 3x^2 - 3x; dx/dy | x=2 Would this turn into 1/3x^2 - 3x and then I plug 2 in and solve? 4. The percentage y (of total personal consumption) an individual spends on food is approximately y = 35x−0.25 percentage points (6.5 ≤ x ≤ 17.5) where x is the percentage the individual spends on education. An individual finds that she is spending x = 7 + 0.2t percent of her personal consumption on education, where t is time in months since January 1. Use direct substitution to express the percentage y as a function of time t (do not simplify the expression). (NOTE: January 1 is represented by t = 0.) y(t) = Use the chain rule to estimate how fast the percentage she spends on food is changing on October 1. (Round your answer to two decimal places.) ______ Would I be using dx/dy and so 7+0.2t / 35x^-0.25 as the equation? So y(t) = 7/35x^-0.25?
Calculus derivative question?
Just dy/dx both sides ^^ Let's start with the easier side first. Derivative of 5x² = 10x Derivative of y(e^x) - (y^3) is a little tricky.. Let's do this term by term y(e^x). This is a basic product rule. You derive the first term, keep the second term, then add the derivative of the second and keep the first term.. (or in notation: derivative of fg = f ' g + g' f ) So the derivative of y(e^x) = y'(e^x) + y(e^x) (y^3). is just a simple power rule. But remember, you have to take the derivative of the base too! Since it's not x, we have to remember to multiply by y' (well techinically it's the chain rule) Derivative of y^3 = 3y² y' So whole thing together.. y'(e^x) + y(e^x) - 3y² y' = 10x Put y(e^x) to right side (because it doesn't have y' as a factor) y'(e^x) - 3y² y' = 10x - y(e^x) Factor out y' y'(e^x - 3y²) = 10x - y(e^x) y' = [ 10x - y(e^x) ] / [ (e^x - 3y²) ] And if you REALLY want, you can try and solve for y in the oringal equation and plug it in so you have y' all in terms of x.. But I don't wanna do that and I'm not if it's possible hehe. Hope I helped
Derivative question please help!?
Because you have a function as an exponent you have to natural log this problem. y=3^cos(x) ln(y) = ln(3^cos(x)) use log rules, when something to a power is logged, it's the same as multiplying ln(y) = ln(cos x) ln(3) now take the derivative, remembering that ln(3) is constant 1/y dy/dx = 1/(cos x) (-sin x) 1/y dy/dx = -sin(x)/cos(x) -sinx/cosx = -tanx 1/y dy/dx = -tanx dy/dx = -tanxy Remember to plug in your original equation for y, as you said y=3^cos(x) dy/dx=(-tan(x))(3^cos(x)) Edit: I humbly concede, his answer is MUCH better
Simple derivative questions?
a rectangular plot of farmland will be bounded on one sie by a river and on the other three sides by a single-strand electric fence. with 800m of wire at your disposal, what is the largest area you can enclose? 1) 80,000 m^2 2) 71,111m^2 3) 40,000m^2 4) 78,000m^2 5) 85,000m^2 an efficiency study of the morning shift at a certain factory indicates that an average worker who arrives on the job at 8 am will have produced Q(t) = -t^3 + 6t^2 + 24t units t hours later. What is the worker's rate of production at 11:00 am? 1) 22 units per hour 2) 24 units per hour 3) 30 units per hour 4) 33 units per hour 5) 35 units per hour Two trucks leave a truck stop at the same time. Truck A travels east at 40 mph and truck B travels north at 30 mph. How fast is the distance between the convoys changing 6 minutes later, when truck A is 4 miles from the truck stop and truck B is 3 miles from the truck stop. 1) 40mph 2) 45mph 3) 50mph 4) -45mph 5) -40mph
AP Calculus Derivative Question?
Chuck, if it can't be punched just do what you know you should - product rule. dy/dx = (0.6^x)[0] + 250[0.6^x(ln0.6)] and drop the first term. You need to know what the derivative of a constant to a variable power is. That's memorization. :)
Intuitive explanation of derivatives
Quick partial derivative question?
that's basically an utility of the quotient rule. basically manage one variable consistent and diff, 2 circumstances on the various. Repeat the technique treating the various variable as a consistent. rather messy yet no longer genuine stressful. a solid try problem!!!!
Quick second derivative question?
you have some errors on your solutions for f' and f''. first of all, you're ultimate that d/dx [5sinx] = 5cosx, yet you're forgetting that d/dx [cos3x] = -3sin3x, by way of fact we would desire to apply the chain rule right here. The spinoff of the outer function evaluated on the indoors function is -sin3x, yet then we would desire to multiply this by utilising the spinoff of the indoors function to get -3sin3x. to that end, f '(x)=5cosx - 3sin3x. additionally, f ''(x) = -5sinx - 9cos3x. keep in mind that the spinoff of cosx is -sinx, providing you with the 1st area of f'', and likewise which you're able to desire to apply the chain rule back right here to tug out yet another ingredient of three. to that end,: f '(x)=5cosx - 3sin3x f ''(x)= -5sinx - 9cos3x
Easy derivative question just want to make sure I did it right?
As long as you are taking the derivative with respect to X (d/dx). The first derivative with respect to x is 4x, you are correct.
Trigonometry and derivative question?
Function(x)= senxcosx if <= pi/2 and x/(x-(pi/2)) if x >pi/2 1.1) show f(x)=(sen(2x))/2 , A(upside down) x E(sign looks like E) ]-infinite, pi/2] 1.2) study the continuity of function when function x=pi/2 1.3) study the vertical asymptote if it exists 1.4) what is derivative of f(x) Guys this is really important, I really do need to know this the more early possible. Thank you for helping. This will make or break my future