To differentiate or calculate the first differential of f(x)='(x) from first principles means to find the first differential of f(x)=f(x) by using the expression Hence d(f(x)) dx f'(x) = - f'(x) = Substitute (x+8x) for x in f(x) = x² to give f'(x) = lim 8x-0 d(f(x)) f(x+8x)-f(x)\ dx Sx f(x+8x)=(x+6x)=x²+2x6x+(8x)² = lim 8x->0 x²+2x8x+(6x)²-x² 8x lim 8x-0 2x8x+(8x)'] Sx au ->> As 8x → 0, [2x + 8x] →→ 2x, so that the differential of x² is 2x. Hence at x =2, the first differential of the gradient of the curve f'(x) = 4. Find lim (2x+8x) dx-0 b. f(x) of f(x) = x³+3x²+5x at x=5.5

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To differentiate or calculate the first differential of f(x)='(x) from first
principles means to find the first differential of f(x)=f(x)
by using the expression
Hence
d(f(x))
dx
f'(x) = -
f'(x) = lim
8x-0
f'(x)= a(f(x))
dx
Substitute (x+8x) for x in f(x) = x² to give
= lim
8x-0
f(x+8x)=(x+8x)=x²+2x5x+(8x)²
[x²+2x8x+(6x)²-x
8x
-*}-
[f(x+8x)-f(x)\
Sx
= lim
8x-0
2x8x+(8x)²
Sx
== lim (2x+8x)
5x-0
As 8x → 0, [2x + 8x] → 2x, so that the differential of x² is 2x.
Hence at x =2, the first differential of the gradient of the curve f'(x) = 4.
Find
Aval
b. f(x) of f(x) = x³+3x²+5x at x=5.5
Transcribed Image Text:To differentiate or calculate the first differential of f(x)='(x) from first principles means to find the first differential of f(x)=f(x) by using the expression Hence d(f(x)) dx f'(x) = - f'(x) = lim 8x-0 f'(x)= a(f(x)) dx Substitute (x+8x) for x in f(x) = x² to give = lim 8x-0 f(x+8x)=(x+8x)=x²+2x5x+(8x)² [x²+2x8x+(6x)²-x 8x -*}- [f(x+8x)-f(x)\ Sx = lim 8x-0 2x8x+(8x)² Sx == lim (2x+8x) 5x-0 As 8x → 0, [2x + 8x] → 2x, so that the differential of x² is 2x. Hence at x =2, the first differential of the gradient of the curve f'(x) = 4. Find Aval b. f(x) of f(x) = x³+3x²+5x at x=5.5
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