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Differential Calculus

Question : 6 of 69

Marks: +1, -0

f(x)=1+2x+3x^{2}+\;\frac{7}{2-x}

f'(x)

Solution:

Concept:

Differentiate each term of

f(x)

using standard rules: constant, power, and chain rule for the fraction.

Explanation:

Given

f(x)=1+2x+3x^{2}+\frac{7}{2-x}

Derivative of

1

0

Derivative of

2x

2

Derivative of

3x^{2}

6x

\frac{7}{2-x}=7(2-x)^{-1}

Using chain rule,

\frac{d}{dx}(2-x)^{-1}=(-1)(2-x)^{-2}\cdot(-1)=(2-x)^{-2}

Thus derivative of

\frac{7}{2-x}

7\cdot(2-x)^{-2}=\frac{7}{(2-x)^{2}}

f'(x)=2+6x+\frac{7}{(2-x)^{2}}

Answer:

2+6x+\frac{7}{(2-x)^{2}}

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