Please help me.
1. Use the definition of the derivative to find f'(x), if f(x)=x^-2
2.Find the derivative of y=3t^5 - 5√t + 7/t
Use the form:
f'(x^n) = nx^(n-1)
1.
f ( x ) = x ^ - 2 = 1 / x ^ 2
f'(x) = lim Δ f ( x ) / Δx
Δx ->0
Δ f ( x ) = 1 / ( x + Δx ) ^ 2 - 1 / x ^ 2 =
[ x ^ 2 - ( x + Δx ) ^ 2 ] / [ ( x + Δx ) ^ 2 * x ^ 2 ] =
[ x ^ 2 - ( x ^ 2 + 2 x * Δx + Δx ^ 2 ) ] / [ ( x + Δx ) ^ 2 * x ^ 2 ] =
( x ^ 2 - x ^ 2 - 2 x * Δx - Δx ^ 2 ) / [ ( x + Δx ) ^ 2 * x ^ 2 ] =
( - 2 x * Δx - Δx ^ 2 ) / [ ( x + Δx ) ^ 2 * x ^ 2 ] =
- Δx * ( 2 x + Δx ) / [ ( x + Δx ) ^ 2 * x ^ 2 ]
Δ f ( x ) / Δx = { - Δx * ( 2 x + Δx ) / [ ( x + Δx ) ^ 2 * x ^ 2 ] } / Δx =
- Δx * ( 2 x + Δx ) / [ Δx * ( x + Δx ) ^ 2 * x ^ 2 ] =
- ( 2 x + Δx ) / [ ( x + Δx ) ^ 2 * x ^ 2 ]
When Δx->0 then:
- ( 2 x + Δx ) = - ( 2 x + 0 ) = - 2 x
( x + Δx ) ^ 2 = ( x + 0 ) ^ 2 = x ^ 2
( x + Δx ) ^ 2 * x ^ 2 = ( x + 0 ) ^ 2 * x ^ 2 =
x ^ 2 * x ^ 2 = x ^ 4
so
f'(x) = lim Δ f ( x ) / Δx =
Δx ->0
lim - ( 2 x + Δx ) / [ ( x + Δx ) ^ 2 * x ^ 2 ] =
Δx ->0
- 2 x / x ^ 4 =
- 2 / x ^ 3
f' ( x ^ - 2 ) = - 2 / x ^ 3
2.
y = 3 t ^ 5 - 5 √t + 7 / t =
3 t ^ 5 - 5 * t ^ ( 1 / 2 ) + 7 * t ^ - 1
Use the power rule : ( t ^ n ) ´ = n * t ^ ( n - 1 )
y ´ = ( 3 t ^ 5 - 5 * t ^ ( 1 / 2 ) + 7 * t ^ - 1 ) ´ =
3 * 5 t ^ ( 5 - 1 ) - 5 * ( 1 / 2 ) * t ( 1 / 2 - 1 ) + 7 * ( - 1 ) * t ^ ( - 1 - 1 ) =
15 t ^ 4 - ( 5 / 2 ) * t ( - 1 / 2 ) - 7 * t ^ - 2 =
15 t ^ 4 - 7 / t ^ 2 - 5 / 2√t
Sure, I'd be happy to help you with your questions!
1. To find the derivative of f(x) = x^-2 using the definition of the derivative, we need to apply the limit definition of the derivative. The definition of the derivative is given by:
f'(x) = lim(h->0) [f(x+h) - f(x)] / h
Substituting f(x) = x^-2 into the formula, we get:
f'(x) = lim(h->0) [(x + h)^-2 - x^-2] / h
To simplify this expression, we need to use algebraic manipulations. First, let's combine the fractions:
f'(x) = lim(h->0) [((x + h)^-2 - x^-2) * (x + h)^2 * x^2] / (h * (x + h)^2 * x^2)
Expanding and simplifying, we get:
f'(x) = lim(h->0) [(x^2 - (x + h)^2) / (h * (x + h)^2 * x^2)]
Next, we can simplify the numerator:
f'(x) = lim(h->0) [(-2xh - h^2) / (h * (x + h)^2 * x^2)]
Now, we can cancel out the h term:
f'(x) = lim(h->0) [(-2x - h) / ((x + h)^2 * x^2)]
Finally, we can take the limit as h approaches 0:
f'(x) = (-2x) / (x^2)^2
Simplifying further, we get:
f'(x) = -2x / x^4
Thus, the derivative of f(x) = x^-2 is f'(x) = -2x / x^4