Asked by Amelia
differentiate from first principles
y=1/x^2
i get to this then i get stuck
f(x+h)-f(x) = 1/(x+h)^2 - 1/x^2
= x^2 - x^2 -2xh -h^2
above divided by(x^2+2xh+h^2)(x^2)
= -2xh -h^2/x^4+2x^3h+h^2x^2
then i know i need to divide by h but i cant seem to get to what i want
y=1/x^2
i get to this then i get stuck
f(x+h)-f(x) = 1/(x+h)^2 - 1/x^2
= x^2 - x^2 -2xh -h^2
above divided by(x^2+2xh+h^2)(x^2)
= -2xh -h^2/x^4+2x^3h+h^2x^2
then i know i need to divide by h but i cant seem to get to what i want
Answers
Answered by
MathMate
You're on the right track. You just have to expand the numerator, without expanding the denominator.
The reason is because in the numerator, you would like to see the x² terms disappear, surfacing the -2xh term which is critical to cancel the h in the denominator. As h→0, (x+h)² would evaluate like x² in the denominator. The complete works would look like:
f(x)=1/x²
Lim x→0 f(x)
=Lim x→0 (f(x+h)-f(x))/h
=Lim x→0 (1/(x+h)²-1/x²)/h
Lim x→0 (x²-x²-2xh-h²)/(h(x+h)²x²))
=Lim x→0 (-2xh-h²)/(h(x+h)²x²))
=Lim x→0 (-2x-h) / (x+h)²x²)
=-2/x³
The reason is because in the numerator, you would like to see the x² terms disappear, surfacing the -2xh term which is critical to cancel the h in the denominator. As h→0, (x+h)² would evaluate like x² in the denominator. The complete works would look like:
f(x)=1/x²
Lim x→0 f(x)
=Lim x→0 (f(x+h)-f(x))/h
=Lim x→0 (1/(x+h)²-1/x²)/h
Lim x→0 (x²-x²-2xh-h²)/(h(x+h)²x²))
=Lim x→0 (-2xh-h²)/(h(x+h)²x²))
=Lim x→0 (-2x-h) / (x+h)²x²)
=-2/x³
Answered by
MathMate
Lim x→0 f(x)
should read
f'(x)
should read
f'(x)
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