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Practice — Maths 11 • Differentiation

Computing derivatives

Difficulty: ★★☆

1804

The function \(f(x)=e^x(x^2+2x+4)\) is given. Find \(f'(0)\), where \(f'(x)\) is the derivative of \(f(x)\).

Your Answer:

Solution:

Use the product rule:

\(f'(x)=(e^x)'(x^2+2x+4)+e^x(x^2+2x+4)'\).

Since \((e^x)'=e^x\) and \((x^2+2x+4)'=2x+2\), we get:

\(f'(x)=e^x(x^2+2x+4)+e^x(2x+2)=e^x(x^2+4x+6)\).

Substitute \(x=0\):

\(f'(0)=e^0(0^2+4\cdot0+6)=1\cdot6=6\).

Computing derivatives

Difficulty: ★★☆

1805

The function \(f(x)=(x^2-3x+3)^4\) is given. Find \(f'(1)\), where \(f'(x)\) is the derivative of \(f(x)\).

Your Answer:

-8

-4

Solution:

Use the chain rule:

\(f'(x)=4(x^2-3x+3)^3\cdot(2x-3)\).

Substitute \(x=1\):

\(f'(1)=4(1-3+3)^3\cdot(2-3)=4\cdot1^3\cdot(-1)=-4\).

Computing derivatives

Difficulty: ★★☆

1806

The function \(f(x)=\dfrac{x}{x-3}\) is given. Find \(f'(4)\), where \(f'(x)\) is the derivative of \(f(x)\).

Your Answer:

-1

-3

Solution:

Use the quotient rule:

\(f'(x)=\dfrac{(x-3)\cdot1-x\cdot1}{(x-3)^2}=\dfrac{x-3-x}{(x-3)^2}=\dfrac{-3}{(x-3)^2}\).

Substitute \(x=4\):

\(f'(4)=\dfrac{-3}{(4-3)^2}=-3\).

Computing derivatives

Difficulty: ★★☆

1807

The function \(f(x)=2e^{x^2-2x}\) is given. Find \(f'(2)\), where \(f'(x)\) is the derivative of \(f(x)\).

Your Answer:

Solution:

Use the chain rule:

\(f'(x)=2e^{x^2-2x}\cdot(2x-2)\).

Substitute \(x=2\):

\(f'(2)=2e^{4-4}\cdot(4-2)=2e^0\cdot2=2\cdot1\cdot2=4\).

Computing derivatives

Difficulty: ★★☆

1808

The function \(f(x)=e^x(3x-5)\) is given. Find \(f'(0)\), where \(f'(x)\) is the derivative of \(f(x)\).

Your Answer:

-5

-2

Solution:

Use the product rule:

\(f'(x)=e^x(3x-5)+e^x\cdot3=e^x(3x-2)\).

Substitute \(x=0\):

\(f'(0)=e^0(3\cdot0-2)=1\cdot(-2)=-2\).