Understanding Gradient Functions in Calculus
Explore the concept of gradient functions in calculus, including differentiation, finding gradients of curves, and understanding how gradients change. Learn about the gradient functions of y = x^2 and y = x^3 and how to calculate gradients at specific points. Discover the significance of positive gradients in increasing functions.
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Presentation Transcript
The Gradient Function Understand that the gradient of a curve at a point is the gradient of the tangent Understand the notation for differentiation Differentiate using standard results Find the gradient of a function
The Gradient Function For a straight line, the gradient is constant: ? = 3 However, for a curve the gradient varies. We can no longer have a single value for the gradient; we ideally want an expression in terms of ? that gives us the gradient for any value of ? (unsurprisingly known as the gradient function).
The Gradient Function ? = ? ? = ? ? = ?? ? = ? ? = ? ? = ? ? = ? ? = ? ? -3 -2 -1 0 1 2 3 ? ? ? ? ? ? ? Gradient -6 -4 -2 0 2 4 6 By looking at the relationship between ? and the gradient at that point, can you come up with an expression, in terms of ? for the gradient? ???????? ???????? = ?? ?
The Gradient Function of y = x2 The gradient of y = x2 changes at a constant rate and so the gradient function of y = x2 is linear.
The Gradient Function So the gradient of the tangent to the curve y = x2 at the general point (x, y) is 2x. 2x is often called the gradient function or the derived function of y = x2. If the curve is written using function notation as y = f(x), then the derived function can be written as f (x). So, if: f(x) = x2 Then: f (x) = 2x This notation is useful if we want to find the gradient of f(x) at a particular point. For example, the gradient of f(x) = x2 at the point (5, 25) is: f (5) = 2 5 =10
The Gradient of y = x3 So, the gradient of the tangent to the curve y = x3 at the general point (x, y) is 3x2. So if, f(x) = x3 Then: f (x) = 3x2 What is the gradient of the tangent to the curve f(x) = x3 at the point ( 2, 8)? f (x) = 3x2 f ( 2) = 3( 2)2 = 12 The gradient function of f(x) = x3 is of the form f (x) = 3x2 Whatever value x takes 3x2 will always be positive, so the Explain why the gradient of f(x) = x3 will always be positive. gradient will always be positive. f(x) = x3 is an example of an increasing function.
? ? = ???+ ?, ? ? = ???? 1 For example, if y = 2x4 dy dx 4 1 =2 4 = 3 x 8x Suppose we have a function of the form y = c, where c is a constant number. This corresponds to a horizontal line through the point (0, c). The gradient of this line is always equal to 0. y c If y = c dy dx 0 x =0
If ? = ??? then ?? i.e. multiply by the power and reduce the power by 1 ??= ???? 1 (where ?,? are constants) Examples: ?? ??= 5?4 ? Power is 5, so multiply by 5 then reduce power by 5. ? = ?5 The power need not be an integer! Remember to use ? ? not ?? ? ? =1 1 2 2? 1 ?(?) = ? ? 2 ?? ?? ??= 12?5 ? = 2?6 ? Why would it be incorrect to say that ? = ?? differentiates to ?? The rule only works when the base is ? and the power is a constant. Neither is true here! Note that ??is a power of ? whereas 2? is an exponential term (which you will encounter more at A-level), and therefore differentiate differently. You will learn how to differentiate exponential terms in Year 13. ??= ? ?? ?? ? ?4= ? 3 ? ? = 3? 4 ? ? ? ? = ?? ??= 3?2 ? ? ?6= ?3 ? ? =
Your Turn Find the gradient of the curve y = 3x4 at the point ( 2, 48). dy dx 3 =12 x Differentiating: At the point ( 2, 48) x = 2 so: dy dx 3 =12( 2) =12 8 = 96 The gradient of the curve y = 3x4 at the point ( 2, 48) is 96.
Differentiating Polynomials Suppose we want to differentiate a polynomial function. For example: Differentiate y = x4 + 3x2 5x + 2 with respect to x. dy dx We can differentiate each term in the function to give . dy dx = 4x3+ 6x 5 So: In general, if y is made up of the sum or difference of any given number of functions, its derivative will be made up of the derivative of each of the functions. dy dx If = ( ) ( ) then y f x = '( ) '( ) x g x f g x
Differentiating Polynomials Differentiate ? = ?2+ 4? + 3 First thing to note: If ? = ? ? + ? ? then ?? ??= ? ? + ? ? i.e. differentiate each term individually in a sum/subtraction. ?? ??= 2? + 4 ? ? ? ? = 4? = 4?1 ? = 3 = 3?0 Therefore applying the usual rule: ?? ??= 4?0= 4 Therefore applying the usual rule: ?? ??= 0? 1= 0 Alternatively, if you compare ? = 4? to ? = ?? + ?, it s clear that the gradient is fixed and ? = 4. Alternatively, if you sketch ? = 4, the line is horizontal, so the gradient is 0.
