Tomasz Lechowski Batory 3IB Studies November Derivatives Find the derivative of each of the following functions: f (x

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Differentiation

Tomasz Lechowski Batory 3IB Studies November 9, 2019 1 / 10

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Intro

The presentation will consist of some practice questions on differentiation and tangent lines.

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Derivatives

Find the derivative of each of the following functions:

f (x ) = 2x²− 3x⁵. This one is very simple:

f⁰(x ) = 2 × 2x¹− 3 × 5x⁴ = 4x − 15x⁴

f (x ) = 3x + 1 − 1

x. We first rewrite f (x ) = 3x + 1 − x⁻¹, now we have:

f⁰(x ) = 3 + 0 − (−1)x⁻²= 3 + 1 x² f (x ) = x⁴

2 + 4

x². We rewrite f (x ) = 1

2x⁴+ 4x⁻², we have:

f⁰(x ) = 1

2× 4x³+ 4 × (−2)x⁻³= 2x³− 8 x³

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Derivatives

f⁰(x ) = 2 × 2x¹− 3 × 5x⁴ = 4x − 15x⁴

f (x ) = 3x + 1 − 1

f⁰(x ) = 3 + 0 − (−1)x⁻²= 3 + 1 x² f (x ) = x⁴

2 + 4

f⁰(x ) = 1

2× 4x³+ 4 × (−2)x⁻³= 2x³− 8 x³

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Derivatives

f⁰(x ) = 2 × 2x¹− 3 × 5x⁴ = 4x − 15x⁴

f (x ) = 3x + 1 − 1

f⁰(x ) = 3 + 0 − (−1)x⁻²= 3 + 1 x² f (x ) = x⁴

2 + 4

f⁰(x ) = 1

2× 4x³+ 4 × (−2)x⁻³= 2x³− 8 x³

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Derivatives

f⁰(x ) = 2 × 2x¹− 3 × 5x⁴ = 4x − 15x⁴

f (x ) = 3x + 1 − 1

f⁰(x ) = 3 + 0 − (−1)x⁻²= 3 + 1 x² f (x ) = x⁴

2 + 4

f⁰(x ) = 1

2× 4x³+ 4 × (−2)x⁻³= 2x³− 8 x³

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Derivatives

f⁰(x ) = 2 × 2x¹− 3 × 5x⁴ = 4x − 15x⁴

f (x ) = 3x + 1 − 1

f⁰(x ) = 3 + 0 − (−1)x⁻²= 3 + 1 x² f (x ) = x⁴

2 + 4

f⁰(x ) = 1

2× 4x³+ 4 × (−2)x⁻³= 2x³− 8 x³

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Derivatives

f⁰(x ) = 2 × 2x¹− 3 × 5x⁴ = 4x − 15x⁴

f (x ) = 3x + 1 − 1

x. We first rewrite f (x ) = 3x + 1 − x⁻¹,now we have:

f⁰(x ) = 3 + 0 − (−1)x⁻²= 3 + 1 x² f (x ) = x⁴

2 + 4

f⁰(x ) = 1

2× 4x³+ 4 × (−2)x⁻³= 2x³− 8 x³

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Derivatives

f⁰(x ) = 2 × 2x¹− 3 × 5x⁴ = 4x − 15x⁴

f (x ) = 3x + 1 − 1

f⁰(x ) = 3 + 0 − (−1)x⁻²= 3 + 1 x² f (x ) = x⁴

2 + 4

f⁰(x ) = 1

2× 4x³+ 4 × (−2)x⁻³= 2x³− 8 x³

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Derivatives

f⁰(x ) = 2 × 2x¹− 3 × 5x⁴ = 4x − 15x⁴

f (x ) = 3x + 1 − 1

f⁰(x ) = 3 + 0 − (−1)x⁻²= 3 + 1 x² f (x ) = x⁴

2 + 4

f⁰(x ) = 1

2× 4x³+ 4 × (−2)x⁻³= 2x³− 8 x³

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Derivatives

f⁰(x ) = 2 × 2x¹− 3 × 5x⁴ = 4x − 15x⁴

f (x ) = 3x + 1 − 1

f⁰(x ) = 3 + 0 − (−1)x⁻²= 3 + 1 x² f (x ) = x⁴

2 + 4

x². We rewrite f (x ) =1

f⁰(x ) = 1

2× 4x³+ 4 × (−2)x⁻³= 2x³− 8 x³

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Derivatives

f⁰(x ) = 2 × 2x¹− 3 × 5x⁴ = 4x − 15x⁴

f (x ) = 3x + 1 − 1

f⁰(x ) = 3 + 0 − (−1)x⁻²= 3 + 1 x² f (x ) = x⁴

2 + 4

2x⁴+ 4x⁻²,we have:

f⁰(x ) = 1

2× 4x³+ 4 × (−2)x⁻³= 2x³− 8 x³

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Derivatives

f⁰(x ) = 2 × 2x¹− 3 × 5x⁴ = 4x − 15x⁴

f (x ) = 3x + 1 − 1

f⁰(x ) = 3 + 0 − (−1)x⁻²= 3 + 1 x² f (x ) = x⁴

2 + 4

f⁰(x ) = 1

2× 4x³+ 4 × (−2)x⁻³= 2x³− 8 x³

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Derivatives

f⁰(x ) = 2 × 2x¹− 3 × 5x⁴ = 4x − 15x⁴

f (x ) = 3x + 1 − 1

f⁰(x ) = 3 + 0 − (−1)x⁻²= 3 + 1 x² f (x ) = x⁴

2 + 4

f⁰(x ) = 1

2× 4x³+ 4 × (−2)x⁻³= 2x³− 8 x³

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Derivatives

f (x ) = 1

2x⁴ + 2x + 6. We rewrite f (x ) = 1

2x⁻⁴+ 2x + 6, now we get:

f⁰(x ) = 1

2 × (−4)x⁻⁵+ 2 + 0 = − 2 x⁵ + 2 f (x ) = 2

3x − 3

2x². Rewrite f (x ) = 2

3x⁻¹−3

2x⁻², so we get:

f⁰(x ) = 2

3 × (−1)x⁻²−3

2 × (−2)x⁻³ = − 2 3x² + 3

x³ f (x ) = 5

x + 3x²− 1

3x⁴. We rewrite f (x ) = 5x⁻¹+ 3x²−1

3x⁻⁴, so we have:

f⁰(x ) = 5 × (−1)x⁻²+ 3 × 2x¹−1

3 × (−4)x⁻⁵= − 5

x² + 6x + 4 3x⁵

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Derivatives

f (x ) = 1

2x⁴ + 2x + 6. We rewrite f (x ) = 1

2x⁻⁴+ 2x + 6,now we get:

f⁰(x ) = 1

2 × (−4)x⁻⁵+ 2 + 0 = − 2 x⁵ + 2 f (x ) = 2

3x − 3

3x⁻¹−3

2x⁻², so we get:

f⁰(x ) = 2

3 × (−1)x⁻²−3

2 × (−2)x⁻³ = − 2 3x² + 3

x³ f (x ) = 5

x + 3x²− 1

3x⁴. We rewrite f (x ) = 5x⁻¹+ 3x²−1

f⁰(x ) = 5 × (−1)x⁻²+ 3 × 2x¹−1

3 × (−4)x⁻⁵= − 5

x² + 6x + 4 3x⁵

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Derivatives

f (x ) = 1

2x⁴ + 2x + 6. We rewrite f (x ) = 1

f⁰(x ) = 1

2 × (−4)x⁻⁵+ 2 + 0 = − 2 x⁵ + 2 f (x ) = 2

3x − 3

3x⁻¹−3

2x⁻², so we get:

f⁰(x ) = 2

3 × (−1)x⁻²−3

2 × (−2)x⁻³ = − 2 3x² + 3

x³ f (x ) = 5

x + 3x²− 1

3x⁴. We rewrite f (x ) = 5x⁻¹+ 3x²−1

f⁰(x ) = 5 × (−1)x⁻²+ 3 × 2x¹−1

3 × (−4)x⁻⁵= − 5

x² + 6x + 4 3x⁵

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Derivatives

f (x ) = 1

2x⁴ + 2x + 6. We rewrite f (x ) = 1

f⁰(x ) = 1

2 × (−4)x⁻⁵+ 2 + 0 = − 2 x⁵ + 2 f (x ) = 2

3x − 3

3x⁻¹−3

2x⁻², so we get:

f⁰(x ) = 2

3 × (−1)x⁻²−3

2 × (−2)x⁻³ = − 2 3x² + 3

x³ f (x ) = 5

x + 3x²− 1

3x⁴. We rewrite f (x ) = 5x⁻¹+ 3x²−1

f⁰(x ) = 5 × (−1)x⁻²+ 3 × 2x¹−1

3 × (−4)x⁻⁵= − 5

x² + 6x + 4 3x⁵

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Derivatives

f (x ) = 1

2x⁴ + 2x + 6. We rewrite f (x ) = 1

f⁰(x ) = 1

2 × (−4)x⁻⁵+ 2 + 0 = − 2 x⁵ + 2 f (x ) = 2

3x − 3

3x⁻¹−3

2x⁻², so we get:

f⁰(x ) = 2

3 × (−1)x⁻²−3

2 × (−2)x⁻³ = − 2 3x² + 3

x³ f (x ) = 5

x + 3x²− 1

3x⁴. We rewrite f (x ) = 5x⁻¹+ 3x²−1

f⁰(x ) = 5 × (−1)x⁻²+ 3 × 2x¹−1

3 × (−4)x⁻⁵= − 5

x² + 6x + 4 3x⁵

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Derivatives

f (x ) = 1

2x⁴ + 2x + 6. We rewrite f (x ) = 1

f⁰(x ) = 1

2 × (−4)x⁻⁵+ 2 + 0 = − 2 x⁵ + 2 f (x ) = 2

3x − 3

3x⁻¹−3

2x⁻²,so we get:

f⁰(x ) = 2

3 × (−1)x⁻²−3

2 × (−2)x⁻³ = − 2 3x² + 3

x³ f (x ) = 5

x + 3x²− 1

3x⁴. We rewrite f (x ) = 5x⁻¹+ 3x²−1

f⁰(x ) = 5 × (−1)x⁻²+ 3 × 2x¹−1

3 × (−4)x⁻⁵= − 5

x² + 6x + 4 3x⁵

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Derivatives

f (x ) = 1

2x⁴ + 2x + 6. We rewrite f (x ) = 1

f⁰(x ) = 1

2 × (−4)x⁻⁵+ 2 + 0 = − 2 x⁵ + 2 f (x ) = 2

3x − 3

3x⁻¹−3

2x⁻², so we get:

f⁰(x ) = 2

3 × (−1)x⁻²−3

2 × (−2)x⁻³ = − 2 3x² + 3

x³ f (x ) = 5

x + 3x²− 1

3x⁴. We rewrite f (x ) = 5x⁻¹+ 3x²−1

f⁰(x ) = 5 × (−1)x⁻²+ 3 × 2x¹−1

3 × (−4)x⁻⁵= − 5

x² + 6x + 4 3x⁵

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Derivatives

f (x ) = 1

2x⁴ + 2x + 6. We rewrite f (x ) = 1

f⁰(x ) = 1

2 × (−4)x⁻⁵+ 2 + 0 = − 2 x⁵ + 2 f (x ) = 2

3x − 3

3x⁻¹−3

2x⁻², so we get:

f⁰(x ) = 2

3 × (−1)x⁻²−3

2 × (−2)x⁻³ = − 2 3x² + 3

x³ f (x ) = 5

x + 3x²− 1

3x⁴. We rewrite f (x ) = 5x⁻¹+ 3x²−1

f⁰(x ) = 5 × (−1)x⁻²+ 3 × 2x¹−1

3 × (−4)x⁻⁵= − 5

x² + 6x + 4 3x⁵

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Derivatives

f (x ) = 1

2x⁴ + 2x + 6. We rewrite f (x ) = 1

f⁰(x ) = 1

2 × (−4)x⁻⁵+ 2 + 0 = − 2 x⁵ + 2 f (x ) = 2

3x − 3

3x⁻¹−3

2x⁻², so we get:

f⁰(x ) = 2

3 × (−1)x⁻²−3

2 × (−2)x⁻³ = − 2 3x² + 3

x³ f (x ) = 5

x + 3x²− 1

3x⁴. We rewrite f (x ) = 5x⁻¹+ 3x²−1

f⁰(x ) = 5 × (−1)x⁻²+ 3 × 2x¹−1

3 × (−4)x⁻⁵= − 5

x² + 6x + 4 3x⁵

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Derivatives

f (x ) = 1

2x⁴ + 2x + 6. We rewrite f (x ) = 1

f⁰(x ) = 1

2 × (−4)x⁻⁵+ 2 + 0 = − 2 x⁵ + 2 f (x ) = 2

3x − 3

3x⁻¹−3

2x⁻², so we get:

f⁰(x ) = 2

3 × (−1)x⁻²−3

2 × (−2)x⁻³ = − 2 3x² + 3

x³ f (x ) = 5

x + 3x²− 1

3x⁴. We rewrite f (x ) = 5x⁻¹+ 3x²−1

3x⁻⁴,so we have:

f⁰(x ) = 5 × (−1)x⁻²+ 3 × 2x¹−1

3 × (−4)x⁻⁵= − 5

x² + 6x + 4 3x⁵

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Derivatives

f (x ) = 1

2x⁴ + 2x + 6. We rewrite f (x ) = 1

f⁰(x ) = 1

2 × (−4)x⁻⁵+ 2 + 0 = − 2 x⁵ + 2 f (x ) = 2

3x − 3

3x⁻¹−3

2x⁻², so we get:

f⁰(x ) = 2

3 × (−1)x⁻²−3

2 × (−2)x⁻³ = − 2 3x² + 3

x³ f (x ) = 5

x + 3x²− 1

3x⁴. We rewrite f (x ) = 5x⁻¹+ 3x²−1

f⁰(x ) = 5 × (−1)x⁻²+ 3 × 2x¹−1

3 × (−4)x⁻⁵= − 5

x² + 6x + 4 3x⁵

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Derivatives

f (x ) = 1

2x⁴ + 2x + 6. We rewrite f (x ) = 1

f⁰(x ) = 1

2 × (−4)x⁻⁵+ 2 + 0 = − 2 x⁵ + 2 f (x ) = 2

3x − 3

3x⁻¹−3

2x⁻², so we get:

f⁰(x ) = 2

3 × (−1)x⁻²−3

2 × (−2)x⁻³ = − 2 3x² + 3

x³ f (x ) = 5

x + 3x²− 1

3x⁴. We rewrite f (x ) = 5x⁻¹+ 3x²−1

f⁰(x ) = 5 × (−1)x⁻²+ 3 × 2x¹−1

3 × (−4)x⁻⁵= − 5

x² + 6x + 4 3x⁵

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Gradients

Find the gradient of f (x ) = 3x²− 5x + 1 at (1, −1). We first find the gradient function (derivative):

f⁰(x ) = 3 × 2x¹− 5 + 0 = 6x − 5

Now we want to find the gradient at (1, −1), so when x = 1. The gradient is

f⁰(1) = 6 × 1 − 5 = 1

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Gradients

f⁰(x ) = 3 × 2x¹− 5 + 0 = 6x − 5

f⁰(1) = 6 × 1 − 5 = 1

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Gradients

f⁰(x ) = 3 × 2x¹− 5 + 0 = 6x − 5

f⁰(1) = 6 × 1 − 5 = 1

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Gradients

f⁰(x ) = 3 × 2x¹− 5 + 0 = 6x − 5

f⁰(1) = 6 × 1 − 5 = 1

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Gradients

f⁰(x ) = 3 × 2x¹− 5 + 0 = 6x − 5

f⁰(1) = 6 × 1 − 5 = 1

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Gradients

Find the gradient of f (x ) = 2x + 1 − 2

x at (2, 4). We want to find the gradient function, so we rewrite f (x ) = 2x + 1 − 2x⁻¹, so we get

f⁰(x ) = 2 + 0 − 2 × (−1)x⁻²= 2 + 2 x²

Now we want to find the gradient at (2, 4), so when x = 2. The gradient is f⁰(2) = 2 + 2

2² = 2.5

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Gradients

f⁰(x ) = 2 + 0 − 2 × (−1)x⁻²= 2 + 2 x²

2² = 2.5

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Gradients

f⁰(x ) = 2 + 0 − 2 × (−1)x⁻²= 2 + 2 x²

2² = 2.5

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Gradients

f⁰(x ) = 2 + 0 − 2 × (−1)x⁻²= 2 + 2 x²

2² = 2.5

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Gradients

f⁰(x ) = 2 + 0 − 2 × (−1)x⁻²= 2 + 2 x²

2² = 2.5

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Gradients

Find the gradient of f (x ) = 1 x³ − 1

x² at (1, 0). We want to find the gradient function, so we first rewrite f (x ) = x⁻³− x⁻², so we get

f⁰(x ) = −3x⁻⁴− (−2)x⁻³= − 3 x⁴ + 2

x³

Now we want to find the gradient at (1, 0), so when x = 1. We substitute x = 1 into the gradient function and we get that the gradient is

f⁰(2) = −3 1⁴ + 2

1³ = −1

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Gradients

f⁰(x ) = −3x⁻⁴− (−2)x⁻³= − 3 x⁴ + 2

x³

f⁰(2) = −3 1⁴ + 2

1³ = −1

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Gradients

f⁰(x ) = −3x⁻⁴− (−2)x⁻³= − 3 x⁴ + 2

x³

f⁰(2) = −3 1⁴ + 2

1³ = −1

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Gradients

f⁰(x ) = −3x⁻⁴− (−2)x⁻³= − 3 x⁴ + 2

x³

f⁰(2) = −3 1⁴ + 2

1³ = −1

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Gradients

f⁰(x ) = −3x⁻⁴− (−2)x⁻³= − 3 x⁴ + 2

x³

f⁰(2) = −3 1⁴ + 2

1³ = −1

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Tangent lines

Find the tangent line to the graph of f (x ) = 1

3x³− 2x at (0, 0). We first need to find the gradient of the function at this point, so we find the gradient function:

f⁰(x ) = 1

3× 3x²− 2 = x²− 2 So the gradient at (0, 0) is

f⁰(0) = 0²− 2 = −2

This means that the gradient of the tangent line will also be −2

(m = −2), so the tangent line will be of the form y = −2x + c. To find c we use the point (0, 0) and we get that c = 0. So the tangent line to the graph of f (x ) = 1

3x³− 2x at (0, 0) will have the equation:

y = −2x

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Tangent lines

f⁰(x ) = 1

f⁰(0) = 0²− 2 = −2

y = −2x

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Tangent lines

f⁰(x ) = 1

f⁰(0) = 0²− 2 = −2

y = −2x

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Tangent lines

f⁰(x ) = 1

f⁰(0) = 0²− 2 = −2

y = −2x

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Tangent lines

f⁰(x ) = 1

f⁰(0) = 0²− 2 = −2

y = −2x

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Tangent lines

f⁰(x ) = 1

f⁰(0) = 0²− 2 = −2

y = −2x