Help with Implicit Differentiation in calculus?
u should know : (xy)' =1* y + xy' and (sinx)' = cosx, (tanx)' = (sec²u)*u' (secu)' = (secu*tanu)u' , (tanx)' =( sec²u)u ' 1.) 3(x^2)y+y^3=sin(xy) 3[2xy + x²y ' ]+3y²y ' =cos(xy)(y+xy ' ) y '=[ycos(xy)-6xy] / [3x²+3y²-xcos(xy)] 2.) x^2+y^3=4xy+y^2 2x + 3y²y ' = 4(y +xy ') + 2yy ' y ' =(4y-2x) / (3y²-4x-2y) 3.) sec(xy)=2x^2-y^2+1 ; sec(xy)tan(xy)(y+xy ')= 4x-2yy ' y '=[4x-y*tan(xy)sec(xy)] / [x*tan(xy)sec(xy) + 2y] 4.) tan(xy)+y^2=x^4+y^4 sec²(xy)(y+xy ')+2yy ' = 4x³ +4y³y ' y ' =[y*sec(xy) - 4x³] / { y' [4y³-2y-xsec²(xy) ]} 5.) y^3-3(x^2)(y^2)+csc(x^2+y^2) This is not an equation . so.... no clue.
Calculus: Implicit differentiation?
The whole point of implicit differentiation is to make the work easier. While standard differentiation is an option, it requires a lot of rearranging and messy work. With implicit differentiation, you just work each term individually. The first term is y^2. You need to take the derivative of this term with respect to y, so you get 2y dy The other term is ln(x). This time take the derivative with respect to x to get 1/x dx When you put them together, you get 2y dy = 1/x dx Now rearrange to get dy/dx = 1/(2xy) then plug in and you get the same answer you gave above, 1/(2e). I hope this helps. I don't have my Mathematical Methods book handy to include proof and such of why this works, and I don't remember off the top of my head. For more information, I recommend "Mathematical Methods in Physical Science" by Mary Boas. Its a very good resource for finding easier ways to solve problems.
Calculus Implicit Differentiation?
1) sin(y) = x d(sin(y))/dx = dx/dx cos(y) * dy/dx = 1 dy/dx = sec(y) d(dy/dx)/dx = d(sec(y))/dx d^2y/dx^2 = sec(y)tan(y) * dy/dx As dy/dx = sec(y): d^2y/dx^2 = sec(y)tan(y) * sec(y) = sec^2(y)tan(y) 2) sin(x^2y^2) = x d(sin(x^2y^2)/dx = dx/dx cos(x^2y^2) * d(x^2y^2)/dx = 1 d(x^2y^2)/dx = sec(x^2y^2) x^2 * d(y^2)/dx + y^2 * d(x^2)/dx = sec(x^2y^2) x^2 * 2y * dy/dx + y^2 * 2x = sec(x^2y^2) 2yx^2 * dy/dx = sec(x^2y^2) - 2xy^2 dy/dx = (sec(x^2y^2) - 2xy^2)/2yx^2 If you want to, you can substitute the value of y in terms of x into the equations, but i doubt if this will be necessary. Hope this helps.
Implicit Differentiation ? Calculus help!?
all right. first off, yes, you do use the derivative. Deriving the equations gives you the answer, but you have to derive them with respect to x. so the derivative of x is 1, but the derivative of y is y prime, or (dy/dx). so. 4xy-3x-11=0 becomes 4(1)(dy/dx)-3=0 or 4dy/dx-3=0. then you get the dy/dx on one side. 4dy/dx=3 dy/dx=3/4 now, for the e equation, the derivative of e^x is (derivative of x)e^x, so that's simple enough, if x is one. if not, like e^2x, your answer is 2e^2x. but the derivative of e^y would be e^(dy/dx). e^y+24-e^2=4x^2+2y^2 becomes e^(dy/dx)=8x+4y*dy/dx not sure exactly where to go from there to isolate the dy/dx. sorry. because 24 and e^2 are both coefficients, not variables, so they become 0. since the y is squared in 2y^2, you end up with 2*2*y*(dy/dx), or just 4y(dy/dx). i may be a bit off on something, feel free to correct. i'm just another calc student procrastinating homework and stopped by to help out :) they really need to add a mathematical notations thing for the math questions on here. later
Calculus - Implicit Differentiation help please!?
i've got self assurance you probably did it without using implicit differentiation. y² = ln x Doing it with implicit differentiation: 2y dy/dx = a million/x dy/dx = a million/(2xy) evaluate at x = e, y = a million dy/dx = a million/(2(e)(a million)) = a million/(2e) spectacular answer, yet your differentiation is using chain rule.
Need Help on a Calculus problem (implicit Differentiation)?
First, we need to get an equation involving f(x) and f'(x); to do this, we can differentiate both sides with respect to x and apply the Chain and Product Rules to get: (d/dx){f(x) + x^2[f(x)]^3} = (d/dx)(10) ==> f'(x) + 2x[f(x)]^3 + 3x^2*f'(x)[f(x)]^2 = 0. Notice that the derivative of [f(x)]^3 is, by the Chain Rule, 3[f(x)]^2 times f(x)'s derivative, f'(x); with this and the Product Rule, the derivative of x^2[f(x)]^3 is 2x[f(x)]^3 + 3x^2*f'(x)[f(x)]^2 = 0. Now, substituting in x = 1 yields: f'(1) + 2(1)[f(1)]^3 + 3(1)f'(1)[f(1)]^2 = 0 ==> f'(1) + 2[f(1)]^3 + 3f'(1)[f(1)]^2 = 0. Substituting in f(1) = 2 and using algebra to solve for f'(1) gives: f'(1) + 2(2)^3 + 3f'(1)(2)^2 = 0 ==> f'(1) + 16 + 12f'(1) = 0 ==> 13f'(1) = -16, by isolating f'(1) ==> f'(1) = -16/13, as required. I hope this helps!
Calculus implicit differentiation homework help?
You don't need implicit differentiation for this one. y = (x^2 + 4)(x - 7y) y = x^3 - 7x^2y + 4x - 28y y + 28y + 7x^2y = x^3 + 4x 29y + 7x^2y = x^3 + 4x y(29 + 7x^2) = (x^3 + 4x) y = (x^3 + 4x) / (29 + 7x^2) Let a = x^3 + 4x, so a' = 3x^2 + 4 Let b = 29 + 7x^2, so b' = 14x So y = a/b, so we use the Quotient Rule: y' = (a'b - ab') / (b^2) y' = ((3x^2 + 4)(29 + 7x^2) - (x^3 + 4x)(14x)) / ((29 + 7x^2)^2) y' = ((87x^2 + 21x^4 + 116 + 28x^2) - (14x^4 + 56x^2)) / ((29 + 7x^2)^2) y' = (87x^2 + 21x^4 + 116 + 28x^2 - 14x^4 - 56x^2) / ((29 + 7x^2)^2) y' = (7x^4 + 59x^2 + 116) / ((29 + 7x^2)^2) If x = 1, then y' = 182/1296 = 91/648 =~ 0.1404
Calculus Implicit Differentiation Question Help!?
I need help with 2 problems for my calculus homework. ***1.) Find dy/dx by implicit differentiation: tan(x/y) = 6x+6y ***2.) The graph of the equation x^2+xy+y^2 = 9 is a slanted ellipse illustrated in the figure posted in this url: http://webwork2.asu.edu/webwork2_course_files/Reynolds_MAT_265_Fall_2010/tmp/gif/Section_2.6-prob11-fig.gif Think of y as a function of x. Differentiating implicitly and solving for y' gives: y' = ___ The ellipse has two horizontal tangents. The upper one has the equation: y = ___ The right most vertical tangent has the equation: x = ___ That tangent touches the ellipse where: y = ___ ***Please show work for both these problems. I would really appreciate the help. Thank you very much for whoever helps me understand these questions!
Calculus Help - Derivatives and Implicit Differentiation?
Wow im actually doing this now in my calculus class 1)x^7+4xy^4=10 7x^6Dx+4x(4y^3)Dy+4y^4Dx=0 D(y)=(7x^6+4y^4) /(16xy^3) 2) Need to use the chain rule twice nu^(n-1)*dy/dx (4x+9x^.5)^.5 .5*(2+4.5x^(-.5))(4x+9x^.5)^(-.5). then multiply it out i usually dont though, my teacher is chill. 3)y=(4-1/x)^-1 dy=-((4-1/x)^-2)/(-x^2) 4)y= sqrt(1-(x^2/25) * -9/2) =(-9/2 + 9x^2/50)^.5 dy=.5( .18x)(-9/2 + 9x^2/50)^-.5 Substitue -7 for x to get m=-.303 Use pt slope formula y=-.303(x+7)+(sqrt216)/5 Know what's ironic im actually answering these problems cause i feel like procrastinating from my homework which is the exact same stuff as this.
Implicit differentiation and derivatives, calculus help?
Ok. The key to implicit differentiation is to realize that when you're deriving with respect to a variable (in this case, x) all other variables should be viewed as separate functions. Watch the first one: 9x^2 y + 2x y^2 = 6 Derive both sides. Each term is a MULTIPLICATION of two variables, so product rule applies. (9(2xy + x^2 y')) + (2(2y^2 + 2xyy')) = 0 Bear in mind that deriving y^2 requires Chain rule. Which is why it comes out to 2yy'. Now just distribute and isolate y' 18xy + 9x^2 y' + 4y^2 + 4xyy' = 0 9x^2 y' + 4xyy' = -18xy - 4y^2 y'(9x^2+4xy) = -18xy - 4y^2 y' = (-18xy - 4y^2)/(9x^2 + 4xy) Now let's try the next one using the same method. 2x^2 + 3y^2 = 9 4x + 6yy' = 0 6yy' = -4x y' = -4x/6y y' = -2x/3y Now for d^2 x/d^2y just derive the result. Use quotient rule here. y'' = -6y + 2xyy' / (3y)^2 Now just apply what we found earlier for y' y'' = (-6y + (2xy(-2x/3y))/9y^2 y'' = (-6y - (4x^2)/3)/ 9y^2 y'' = (-18y -4x^2) / (27y^2) Now for the third one we need chain rule. Watch how I do it's y = (1+(3/x))^3 y' = 3(1+3/x)^2 * -3/x^2 Bear in mind that 3/x = 3*1/x, and d/dx 1/x = d/dx x^-1 = -1/x^2 y' = 3(((1+3/x)^2)/x^2) dx Now since we have two functions dividing each other, we need quotient rule. y'' = 3((2(1+3/x)*3/x^2*x^2) - (2x(1+3/x)^2))/x^4 y'' = 3((2+6/x)*3) - 2x(1+3/x)^2) /x^4 y'' = (18+54/x) - (2x(1+3/x)^2) / x^4 You can continue simplifying if you like from here but this should be fine. y = (tan x - sec x)^-1 You can make this easier by simplifying y = (sin x / cos x - 1/ cos x)^-1 y = ((sin x-1)/cos x)^-1 Now we can use chain rule and make this tougher, or just simplify further and apply quotient rule. y = cos x / (sin x - 1) y' = (-sin^2 x + sin x) - (cos^2 x) /(sin x -1)^2 y' = (-sin^2 x + sin x - (1-sin^2 x)) / (sin x - 1)^2 y' = (-sin^2 x + sin x - 1 +sin^2 x ) / (sin x -1)^2 y' = (sin x -1)/(sin x -1)^2 y' = 1 / (sin x -1)