differentiate (2x-3)^2
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explanation:Chain Rule of Differentiation
Chain Rule of DifferentiationLet f(x) = (g o h)(x) = g(h(x))
Chain Rule of DifferentiationLet f(x) = (g o h)(x) = g(h(x))Let u = h(x)
Chain Rule of DifferentiationLet f(x) = (g o h)(x) = g(h(x))Let u = h(x)Using the above, function f may be written as:
Chain Rule of DifferentiationLet f(x) = (g o h)(x) = g(h(x))Let u = h(x)Using the above, function f may be written as:f(x) = g(u)
Chain Rule of DifferentiationLet f(x) = (g o h)(x) = g(h(x))Let u = h(x)Using the above, function f may be written as:f(x) = g(u)the derivative of f with respect to x, f ' is given by:
Chain Rule of DifferentiationLet f(x) = (g o h)(x) = g(h(x))Let u = h(x)Using the above, function f may be written as:f(x) = g(u)the derivative of f with respect to x, f ' is given by:f '(x) = (df / du) (du / dx)
example
f(x) = 4 cos (5x - 2)
f(x) = 4 cos (5x - 2)Solution to Example 1
f(x) = 4 cos (5x - 2)Solution to Example 1Let u = 5x - 2 and f(u) = 4 cos u, hence
f(x) = 4 cos (5x - 2)Solution to Example 1Let u = 5x - 2 and f(u) = 4 cos u, hencedu / dx = 5 and df / du = - 4 sin u
f(x) = 4 cos (5x - 2)Solution to Example 1Let u = 5x - 2 and f(u) = 4 cos u, hencedu / dx = 5 and df / du = - 4 sin uWe now use the chain rule
f(x) = 4 cos (5x - 2)Solution to Example 1Let u = 5x - 2 and f(u) = 4 cos u, hencedu / dx = 5 and df / du = - 4 sin uWe now use the chain rulef '(x) = (df / du) (du / dx) = - 4 sin (u) (5)
f(x) = 4 cos (5x - 2)Solution to Example 1Let u = 5x - 2 and f(u) = 4 cos u, hencedu / dx = 5 and df / du = - 4 sin uWe now use the chain rulef '(x) = (df / du) (du / dx) = - 4 sin (u) (5)We now substitute u = 5x - 2 in sin (u) above to obtain
f(x) = 4 cos (5x - 2)Solution to Example 1Let u = 5x - 2 and f(u) = 4 cos u, hencedu / dx = 5 and df / du = - 4 sin uWe now use the chain rulef '(x) = (df / du) (du / dx) = - 4 sin (u) (5)We now substitute u = 5x - 2 in sin (u) above to obtainf '(x) = - 20 sin (5x - 2)
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