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Implicit Differentiation

This is a simple guide about how to do implicit differentiation Explained in a simple way, if there is a function that is not in the form y = some expression with only x. There are many instances where y cannot be solved explicitly in terms of x, so implicit differentiation required. Other times, it is much faster and “cleaner” to use implicit differentiation For example, given the equation x^2 + y^2 =49, implicit differentiation is the easiest way to find dy/dx To do implicit differentiation, know that the derivative of y is dy/dx. In other words simply add a dy/dx after every y when the derivative is taken. Essentially, this is a continuation of the chain rule. Then, solve for dy/dx in terms of x and y Steps d/dx the entire equation Differentiate x as normal When differentiating y, add a dy/dx afterwards Solve for dy/dx in terms of x and y Example Problems x^2 + y^2 = 49 d/dx(x^2 + y^2) = 49 2x + 2y(dy/dx) = 0 Solve for dy/dx so dy/dx
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Product Rule

This is a simple guide about how to take the derivative of a product of two functions This post will use the notation f '(x) to show the first derivative of f with respect to x To find the derivative of a product, simply use the rule and formula shown below. Given a function h(x) = f(x)*g(x): The derivative h'(x) = f(x)*g'(x) + g(x)*f '(x) Examples Y = xsinx Recall that h'(x) = f(x)*g'(x) + g(x)*f '(x) In this case, f(x) = x and g(x) = sinx Therefore f ‘(x) = 1 and g’(x) = cosx Substituting into the product rule formula, the answer dy/dx = x*cosx + 1*sinx dy/dx = xcosx + sinx Y = (x^3)lnx Recall that h'(x) = f(x)*g'(x) + g(x)*f '(x) In this case, f(x) = x^3 and g(x) = lnx Therefore f ‘(x) = 3x^2 and g’(x) = 1/x Substituting into the formula the answer dy/dx = (x^3)*(1/x) + (lnx)*(3x^2) dy/dx = x^2 + (3x^2)lnx Y = (sinx)(cosx) Recall that h'(x) = f(x)*g'(x) + g(x)*f 
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Graphing Quadratics

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This is a simple guide to graphing quadratics A quadratic equation is most often given in one of two forms as shown below. Standard form:  y = ax^2 + bx + c Vertex form:  y=a(x-h)^2 + k For the standard form ( y = ax^2 + bx + c) follow the steps below to graph Find the vertex Do -b/2a to find the x-coordinate of the vertex Plug in the value from -b/2a into the original equation to solve for the y-coordinate of the vertex Graph the vertex Graph a point before and after the vertex Choose an x-coordinate before the x-coordinate of the vertex Plug this value into the original equation to find the value of the y-coordinate Graph the point Repeat a-c with an x-coordinate after the x-coordinate of the vertex For the second form ( y=a(x-h)^2 + k) follow the steps below to graph Find the vertex The vertex is (h,k) Graph the vertex See step 2 from graphing a quadratic equation in standard form An example problem is y=x^2-4x-5
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