Cho hàm số \(f\left( x \right) = \frac{{x - 3}}{{x + 4}}\); g(x) = xcos2x.

a) \(f'\left( 2 \right) + g'\left( 0 \right) = \frac{{16}}{9}\).

b) Hàm số f(x) có đạo hàm là \(f'\left( x \right) = \frac{1}{{{{\left( {x + 4} \right)}^2}}}\).

c) \(2{\left( {f'\left( x \right)} \right)^2} = \left( {f\left( x \right) - 1} \right)f''\left( x \right)\).

d) Hàm số g(x) có đạo hàm là g'(x) = cos2x + 2xsin2x.

Ta có \(f'\left( x \right) = \frac{7}{{{{\left( {x + 4} \right)}^2}}}\); \(f''\left( x \right) = \frac{{ - 14}}{{{{\left( {x + 4} \right)}^3}}}\); g'(x) = cos2x – 2xsin2x.

a) \(f'\left( 2 \right) = \frac{7}{{{{\left( {2 + 4} \right)}^2}}} = \frac{7}{{36}}\); g'(0) = 1.

Do đó \(f'\left( 2 \right) + g'\left( 0 \right) = \frac{7}{{36}} + 1 = \frac{{43}}{{36}}\).

b) \(f'\left( x \right) = \frac{7}{{{{\left( {x + 4} \right)}^2}}}\).

c) \(2{\left( {f'\left( x \right)} \right)^2} = 2{\left( {\frac{7}{{{{\left( {x + 4} \right)}^2}}}} \right)^2} = \frac{{98}}{{{{\left( {x + 4} \right)}^4}}}\); \(\left( {f\left( x \right) - 1} \right)f''\left( x \right) = \left( {\frac{{ - 7}}{{x + 4}}} \right).\frac{{ - 14}}{{{{\left( {x + 4} \right)}^3}}} = \frac{{98}}{{{{\left( {x + 4} \right)}^4}}}\).

Do đó \(2{\left( {f'\left( x \right)} \right)^2} = \left( {f\left( x \right) - 1} \right)f''\left( x \right)\).

d) g'(x) = cos2x – 2xsin2x.

Đáp án: a) Sai;   b) Sai;   c) Đúng;   d) Sai.