Implicit Differentiation Calculator Implicit differentiation is a special case of the well-known chain rule for derivatives. The majority of differentiation problems in first-year calculus involve functions y written explicitly as functions of x. It is an online tool to do differentiation of implicit function. It makes calculation easy and fun. If any function in implicit form is given then it can easily do the differentiation of that function. Step by Step Calculations Step 1 : Don't solve for y. Step 2 : Now, differentiate each terms on both sides of the equation by using chain rule. Step 3 : Solve for dy/dx. Know More About Free Tutoring Online
Example Problems Find the implicit derivation of the function: y3 - x2 = -10? Step 1 : Given function: y3 - x2 = -10 Step 2 : Possible derivation: ddx (y(x)3 - x2 ) = ddx (-10) Differentiate the sum term by term and factor out constants: => ddx (y(x)3 ) - ddx (x2) = ddx (-10) Use the chain rule, ddx (y(x)3) = du3du dudx , where u = y(x) and du3du = 3u2 = 3 y(x)2 y′(x) - 2x = ddx (-10) The derivative of -10 is zero: = 3 y(x)2 y′(x) - 2x Step 3 : y′(x) = 2x3y2 Answer : dydx = 2x3y2 Find the implicit derivation of the function: x2 + y2 = 25? Step 1 : Given function: x2 + y2 = 25 Step 2 : Possible derivation: ddx (x2 + y(x)2 ) = ddx (25) Differentiate the sum term by term and factor out constants: Learn More :- Free Tutoring
=> ddx ((x)2 ) + ddx (y(x)2) = ddx (25) Use the chain rule, ddx (y(x)2) = du2du dudx , where u = y(x) and du2du = 2u => 2 y(x) (ddx (y(x))) + 2x = ddx (25) The derivative of y(x) is y′(x) + 2x = ddx (25) The derivative of 25 is zero: => 2 y(x) y′(x) + 2x = 0 => dydx = −xy Step 3 : y′(x) = 2x3y2 Answer : dydx = −xy
Graphing Calculator Graphing Calculator is a tool which makes calculations easy and fun. It is used to plot an equations and inequality. Try our Graphing Calculator and get your problems solved instantly. Step by Step CalculationsBack to Top Step 1 : Choose different values for "x" zero and solve for y to get different coordinates. Step 2 : Mark all the coordinates in the graph and join it. Example Problems Graph y = 3x + 5 Step 1 : Given equation: y = 3x + 5 At x = 0 => y = 3x + 5 => y = 3(0) + 5
=> y = 5 At x = 1 => y = 3x + 5 => y = 3(1) + 5 => y = 3 + 5 => y = 8 At x = -2 => y = 3(-2) + 5 => y = -6 + 5 => y = -1 At x = -3 => y = 3(-3) + 5 => y = -9 + 5 => y = -4 So, the coordinates are: (0,5), (1, 8), (-2, -1) and (-3, -4) Graph y= 2x2 - 6x + 4 Step 1 : Given equation: y= 2x2 - 6x + 4 At x = 0 => y = 2x2 - 6x + 4 => y= 2(0)2 - 6(0) + 4 Read More About Free Online Tutor
=> y= 4 At x = 1 => y= 2x2 - 6x + 4 => y= 2(1)2 - 6(1) + 4 => y= 0 At x = -1 => y= 2(-1)2 - 6(-1) + 4 => y= 2 + 6 + 4 => y = 12 At x= 2 => y= 2(2)2 - 6(2) + 4 => y= 8 - 12 + 4 => y = 0 So, the coordinates are: (0, 4), (1, 0), (-1, 12) and (2, 0)
Thank You
TutorVista.com