Determine the derivative of the following: 1.1 f(x) = 5x³ − 4x² + 3x − 8 1.2 y = 3/x² − 4sqrt(x) + 3(x + 2)

Question

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to the calculus questions. 1.1 Determine the derivative of f(x) = 5x³ − 4x² + 3x − 8. Step 1: Apply the power rule for differentiation, d/dx(axⁿ) = anxⁿ⁻¹, and the rule for constants, d/dx(c) = 0. f'(x) = d/dx(5x³) − d/dx(4x²) + d/dx(3x) − d/dx(8) Step 2: Differentiate each term. f'(x) = 5(3)x³⁻¹ − 4(2)x²⁻¹ + 3(1)x¹⁻¹ − 0; f'(x) = 15x² − 8x + 3; f'(x) = 15x² − 8x + 3; 1.2 Determine the derivative of y = (3)/(x²) − 4√(x) + 3(x + 2). Step 1: Rewrite the terms in the form axⁿ. y = 3x⁻² − 4x^(1/2) + 3x + 6 Step 2: Apply the power rule for differentiation to each term. dy/dx = d/dx(3x⁻²) − d/dx(4x^(1/2)) + d/dx(3x) + d/dx(6); dy/dx = 3(−2)x⁻²⁻¹ − 4(1/2)x^(1/2−1) + 3(1)x¹⁻¹ + 0 Step 3: Simplify the expression. dy/dx = −6x⁻³ − 2x^(−1/2) + 3; dy/dx = −(6)/(x³) − (2)/(√(x)) + 3; dy/dx = −(6)/(x³) − (2)/(√(x)) + 3; 2.1 Determine the equation of the tangent passing through x = 2 for g(x) = −3x² + 2x − 1. Step 1: Find the y-coordinate of the point of tangency. Substitute x = 2 into g(x): g(2) = −3(2)² + 2(2) − 1; g(2) = −3(4) + 4 − 1; g(2) = −12 + 4 − 1; g(2) = −9 The point of tangency is (2, −9). Step 2: Find the derivative of g(x) to get the gradient function. g'(x) = d/dx(−3x² + 2x − 1); g'(x) = −3(2)x²⁻¹ + 2(1)x¹⁻¹ − 0; g'(x) = −6x + 2; Step 3: Calculate the gradient of the tangent at x = 2. Substitute x = 2 into g'(x): m = g'(2) = −6(2) + 2; m = −12 + 2; m = −10; Step 4: Use the point-slope form y − y₁ = m(x − x₁) to find the equation of the tangent. Using (x₁, y₁) = (2, −9) and m = −10: y − (−9) = −10(x − 2); y + 9 = −10x + 20; y = −10x + 20 − 9; y = −10x + 11; y = −10x + 11; 2.2 Is the graph of g(x) = −3x² + 2x − 1 increasing or decreasing at x = −5? Give a reason for your answer. Step 1: Use the derivative g'(x) = −6x + 2 (from 2.1). Step 2: Evaluate the derivative at x = −5. g'(−5) = −6(−5) + 2; g'(−5) = 30 + 2; g'(−5) = 32 Step 3: Determine if the graph is increasing or decreasing. Since g'(−5) = 32 > 0, the graph is increasing at x = −5. The graph is increasing at x = −5 because g'(−5) > 0. 3. Determine the coordinate(s) of the stationary point(s) of h(x) = 5x² + 3x − 2. Step 1: Find the derivative of h(x). h'(x) = d/dx(5x² + 3x − 2); h'(x) = 5(2)x²⁻¹ + 3(1)x¹⁻¹ − 0; h'(x) = 10x + 3 Step 2: Set the derivative equal to zero to find the x-coordinate(s) of the stationary point(s). 10x + 3 = 0; 10x = −3; x = −3/10 Step 3: Substitute the x-coordinate back into the original function h(x) to find the y-coordinate. h(−3/10) = 5(−3/10)² + 3(−3/10) − 2; h(−3/10) = 5(9/100) − 9/10 − 2; h(−3/10) = 45/100 − 90/100 − 200/100; h(−3/10) = (45 − 90 − 200)/(100); h(−3/10) = (−245)/(100); h(−3/10) = −49/20 The coordinate of the stationary point is (−3/10, −49/20). (−3/10, −49/20) 4. Given p(x) = 5x². Using first principles determine p'(x). Step 1: Write down the definition of the derivative from first principles. p'(x) = lim_(h → 0) (p(x+h) − p(x))/(h) Step 2: Substitute p(x+h) and p(x) into the formula. p(x+h) = 5(x+h)² = 5(x² + 2xh + h²) = 5x² + 10xh + 5h²; p'(x) = lim_(h → 0) ((5x² + 10xh + 5h²) − 5x²)/(h) Step 3: Simplify the numerator. p'(x) = lim_(h → 0) (10xh + 5h²)/(h) Step 4: Factor out h from the numerator and cancel it with the denominator. p'(x) = lim_(h → 0) (h(10x + 5h))/(h); p'(x) = lim_(h → 0) (10x + 5h) Step 5: Apply the limit as h → 0. p'(x) = 10x + 5(0); p'(x) = 10x; p'(x) = 10x

Determine the derivative of the following: 1.1 f(x) = 5x³ − 4x² + 3x − 8 1.2 y = 3/x² - 4sqrt(x) + 3(x + 2)

Answers

Step-by-step solution

Part 1.1· 2 steps

Apply the power rule for differentiation, $\frac{d}{dx}(ax^n) = anx^{n-1}$ , and the rule for constants, $\frac{d}{dx}(c) = 0$ .

Differentiate each term.

Part 1.2· 3 steps
$\frac{dy}{dx} = -\frac{6}{x^3} - \frac{2}{\sqrt{x}} + 3$

Rewrite the terms in the form $ax^n$ .

Apply the power rule for differentiation to each term.

Simplify the expression.

Part 2.1· 4 steps
$y = -10x + 11$

Find the y-coordinate of the point of tangency.

Find the derivative of $g(x)$ to get the gradient function.

Calculate the gradient of the tangent at $x = 2$ .

Use the point-slope form $y - y_1 = m(x - x_1)$ to find the equation of the tangent.

Part 2.2· 11 steps
$\text{The graph is increasing at } x = -5 \text{ because } g'(-5) > 0$

Use the derivative $g'(x) = -6x + 2$ (from 2.1).

Evaluate the derivative at $x = -5$ .

Determine if the graph is increasing or decreasing.

Find the derivative of $h(x)$ .

Set the derivative equal to zero to find the x-coordinate(s) of the stationary point(s).

Substitute the x-coordinate back into the original function $h(x)$ to find the y-coordinate.

Write down the definition of the derivative from first principles.

Substitute $p(x+h)$ and $p(x)$ into the formula.

Simplify the numerator.

Factor out $h$ from the numerator and cancel it with the denominator.

Apply the limit as $h \to 0$ .

Try one yourself

Still stuck on this one?

More Mathematics Questions

Determine the derivative of the following: 1.1 f(x) = 5x³ − 4x² + 3x − 8 1.2 y = 3/x² - 4sqrt(x) + 3(x + 2)

Answers

Step-by-step solution

Part 1.1· 2 steps

Apply the power rule for differentiation, ddx(axn)=anxn−1\frac{d}{dx}(ax^n) = anx^{n-1}dxd​(axn)=anxn−1, and the rule for constants, ddx(c)=0\frac{d}{dx}(c) = 0dxd​(c)=0.

Differentiate each term.

Part 1.2· 3 stepsdydx=−6x3−2x+3\frac{dy}{dx} = -\frac{6}{x^3} - \frac{2}{\sqrt{x}} + 3dxdy​=−x36​−x​2​+3

Rewrite the terms in the form axnax^naxn.

Apply the power rule for differentiation to each term.

Simplify the expression.

Part 2.1· 4 stepsy=−10x+11y = -10x + 11y=−10x+11

Find the y-coordinate of the point of tangency.

Find the derivative of g(x)g(x)g(x) to get the gradient function.

Calculate the gradient of the tangent at x=2x = 2x=2.

Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​) to find the equation of the tangent.

Part 2.2· 11 stepsThe graph is increasing at x=−5 because g′(−5)>0\text{The graph is increasing at } x = -5 \text{ because } g'(-5) > 0The graph is increasing at x=−5 because g′(−5)>0

Use the derivative g′(x)=−6x+2g'(x) = -6x + 2g′(x)=−6x+2 (from 2.1).

Evaluate the derivative at x=−5x = -5x=−5.