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  • What is the derivative of f (t) = (t^3e^ (1-t) , tan^2t ) ? | Socratic
    What is the derivative of f (t) = (t3e1−t, tan2 t)? Calculus Parametric Functions Derivative of Parametric Functions
  • Special Limits Involving sin (x), x, and tan (x) - Socratic
    Questions and Videos on Special Limits Involving sin(x), x, and tan(x), within Calculus
  • Answers created by Jacob F. - Socratic
    How do you find the equation of a normal line to a curve at a given point? How do I find the integral #int (x*cos (5x))dx# ? How do you find the derivative of #y=x^tan (x)#? How do you find the derivative of #y=tan^2 (3x)#? How do you find the 50th derivative of #y=cos (x)# ? What is notation for the Second Derivative?
  • y=tan^2 (x) Find dy dxThe answer is 2tan (x)sec^2 (x). It . . . - Socratic
    d du (u) = 1, but since d dx (u) differentiates with respect to x, not u, the derivative isn't 1 In terms of your problem, it's more helpful to write d dx tan(x) because it's clear that the derivative is sec2(x)
  • Differentiating Inverse Trigonometric Functions - Socratic
    Questions and Videos on Differentiating Inverse Trigonometric Functions, within Calculus
  • Question #233fc - Socratic
    f (x)=tan (x) Using the limit definition of the derivative, which states that the derivative of f (x) is given by f' (x)=lim_ (Deltaxrarr0) (f (x+Deltax)-f (x
  • Question #ea7ac - Socratic
    d dx(tan^-1(x))=1 (x^2+1) We will use implicit differentiation y=tan^-1(x) We can take tan on both sides (since it'll eliminate the inverse tan, and we already know how to differentiate it): tan(y)=cancel(tan)(cancel(tan^-1)(x)) tan(y)=x Now we can differentiate both sides with respect to x: d dx(tan(y))=dx dx To work out the left hand side we need to be careful Since we're differentiating
  • Question #7b4a1 - Socratic
    Explanation: First, recall the formula: #d dx (tan^-1 (x)) = 1 (1+x^2)# We will need chain rule for our derivative: #d dx (tan^-1 (x+1 x))# #= d (d (x+1 x)) (tan^-1
  • What is the derivative of arctan (x-1)? | Socratic
    Drawing a right triangle with an angle y such that tan(y) = x − 1, we can find that sec(y) = √x2 − 2x + 2 Plugging that in, we get our result: d dx arctan(x − 1) = 1 x2 − 2x +2 Note that the same process shows that the formula for d dx arctan(x) = 1 x2 + 1 If we already have that formula, then we can simply use the chain rule:





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