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Tuesday, May 23, 2017
Derivative of a composite function
Derivation of a composite function
Let's consider a function g. The image by g of any element x of its domain is g(x). Let's consider another function f. The image of g(x) by f is f[g(x)] also written as fog(x). The function fog is called the composed function of g and f.
If g is differentiable for any element x and f is differentiable at g(x) fog(x) = f[g(x)] is differentiable at x. The derivative of the function fog is (fog)'(x) = f'[g(x)].g'(x), The demonstration of this formula is not done here.
The derivative of fog is the product of the derivative of fog by the derivative of g.
If u is a function of x and f is a function of u then f(u) is a composite function. By applying the rule above the derivative of f(u) or f'(u) is equal to the derivative of f(u) multiplied by the derivative of u. We write [f(u)]' = f''(u).u'. If we introduce the notation (d) of differentiability we can write d/dx[f(u)] =d/du[f(u)].du/dx.
In practical applications we have a function f to differentiate with respect to x. We then introduce a function u that is a function of x. Now we have the composite function f(u). The differentiation or derivative of f with respect to x is equal to the derivative of f with respect to u multiplied by the derivative of u with respect to x . This derivative is called the chain rule. There is a chain of operations to do. First we introduce a new function u. Then we calculate the derivative of f as the the composite function f(u) by applying the formula for the derivative of a composite function.
The chain rule holds also the application of the power rule when we work with a complex function with exponents.
The power rule applies by introducing the new function u.
Example 1
Let's calculate the derivative of the function f(x) = (2x+1)²
To make the computation of the derivative easy we introduce the function u. Then the function f becomes f(x) = u². The derivative of the function f with respect to x is the derivative of the expression u² with respect to x . We write d/dx[f(x)] = d/dx[u²]
By applying the formula for the derivative of a composed function we have d/dx[f(x)] = d/du(u²).du/dx.
By calculating d/du(u²) we obtain d/dx[f(x)] = 2u. u'
Let's substitute u: d/dx[f(x)] = 2 (2x+1)(2x+1)'
By calculating the derivative of 2x+1 we obtain d/dx[f(x)] = 2(2x+1)(2) = 4(2x+1) = 8x+1
Example 2
Calulate the derivative of f(x) = (x²+3x+4)²
Let's write u = x^2+3x+4
d/dx[f(x)] = d/dx(x²+3x+4)²
= d/dx(u²)
= d/du(u²).du/dx (Applying the formula of the derivative of a composite function)
= 2u.u'
= 2(x²+3x+4)(x²+3x+4)' (Substituting u)
= 2(x²+3x+4)(2x+3)
= 2(2x³+6x²+8x+3x²+9x+12)
= 2(2x³+9x²+17x+12)
= 4x³+18x²+34x+24)
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Labels:
Calculus,
derivative,
math learning
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