Question

(1) Find the derivative of $f(x) = 4x^3 + 3x$ from first principles (2) Differentiate using power rule $ax^n$ $f(x) = x^3 + x^2 + 3$ $f(x) = -3x^2 - 6x$ at $x = 1$. (a) (b) (3) Differentiate using product rule $y = 2x^2 (3x - 4)$ $f(t) = (4t^2 - t)(t^3 - 8t^2 + 12)$ (a) (b) (4) Differentiate using Chain rule $f(x) = (x^5 + 4)^5$ $f(x) = \sqrt{x^2 - 1}$ (a) (b) (5) Differentiate using Quotient rule $\frac{\sqrt{x} + 2x}{7x - 4x^2}$ $f(x) = \frac{3x^2 - 5x}{2x}$ (a) (b)

          (1) Find the derivative of $f(x) = 4x^3 + 3x$ from first principles
(2) Differentiate using power rule $ax^n$
$f(x) = x^3 + x^2 + 3$
$f(x) = -3x^2 - 6x$ at $x = 1$.
(a)
(b)
(3) Differentiate using product rule
$y = 2x^2 (3x - 4)$
$f(t) = (4t^2 - t)(t^3 - 8t^2 + 12)$
(a)
(b)
(4) Differentiate using Chain rule
$f(x) = (x^5 + 4)^5$
$f(x) = \sqrt{x^2 - 1}$
(a)
(b)
(5) Differentiate using Quotient rule
$\frac{\sqrt{x} + 2x}{7x - 4x^2}$
$f(x) = \frac{3x^2 - 5x}{2x}$
(a)
(b)

(1) Find the derivative of f(x) = 4x^3 + 3x from first principles
(2) Differentiate using power rule ax^n
f(x) = x^3 + x^2 + 3
f(x) = -3x^2 - 6x at x = 1.
(a)
(b)
(3) Differentiate using product rule
y = 2x^2 (3x - 4)
f(t) = (4t^2 - t)(t^3 - 8t^2 + 12)
(a)
(b)
(4) Differentiate using Chain rule
f(x) = (x^5 + 4)^5
f(x) = √(x^2 - 1)
(a)
(b)
(5) Differentiate using Quotient rule
(√(x) + 2x)/(7x - 4x^2)
f(x) = (3x^2 - 5x)/(2x)
(a)
(b)

Added by Ryan M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ahmet Yavas

05/26/2022

Step 1

Step 1: Apply the product rule to differentiate y=23x-4 The product rule states that if y = u*v, then y' = u'v + uv' Let u = 23x and v = -4 Then u' = 23 and v' = 0 So, y' = 23*0 + 23*(-4) = -92 Show more…

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Thanks for your feedback!

f(=3+2 +3 (a) f(=-32-6 at=1. (b) 3 Differentiate using product rule y=23x-4 (a) ft=42-tt3-8t2+12 (b 4) Differentiate using Chain rule fx=(+4s (a) fx=x-1 (b) 5 Differentiate using Quotient rule Vx+2x 7x-4x2 3x2-5x fx= 2x f(a (a) (b)