Đáp án:
$a) f'(x)=x^4-x^3+4x-1\\
b) f'(x)=\dfrac{7}{(x+3)^2}\\
c) f'(x)=6x^2+14x+13\\
2)
a) f'(x)=\dfrac{2x^2\cos x^2-\sin x^2}{x^2}\\
b) g'(x)=x\sqrt{x^2+1}+(x-2)\dfrac{x}{\sqrt{x^2+1}}\\
c) h'(x)=10(x^3-x^2+1)^{9}.(3x^2-2x)\\$
Giải thích các bước giải:
$a) f(x)=\dfrac{x^5}{5}-\dfrac{x^4}{4}+2x^2-x+7\\
\Rightarrow f'(x)=\left (\dfrac{x^5}{5}-\dfrac{x^4}{4}+2x^2-x+7 \right )'\\
=\dfrac{5}{5}x^4-\dfrac{4}{4}x^3+2.2x-1\\
=x^4-x^3+4x-1\\
b) f(x)=\dfrac{2x-1}{x+3}\\
\Rightarrow f'(x)=\left (\dfrac{2x-1}{x+3} \right )'\\
=\dfrac{(2x-1)'(x+3)-(2x-1)(x+3)'}{(x+3)^2}\\
=\dfrac{2(x+3)-(2x-1).1}{(x+3)^2}\\
=\dfrac{2x+6-2x+1}{(x+3)^2}\\
=\dfrac{7}{(x+3)^2}\\
c) f(x)=(2x+1)(x^2+3x+5)\\
\Rightarrow f'(x)=\left [(2x+1)(x^2+3x+5) \right ]'\\
=(2x+1)'(x^2+3x+5)+(2x+1)(x^2+3x+5)'\\
=2(x^2+3x+5)+(2x+1)(2x+3)\\
=2x^2+6x+10+4x^2+6x+2x+3\\
=6x^2+14x+13\\
2)
a) f(x)=\dfrac{\sin x^2}{x}\\
\Rightarrow f'(x)=\left (\dfrac{\sin x^2}{x} \right )'\\
=\dfrac{(\sin x^2)'x-\sin x^2.x'}{x^2}\\
=\dfrac{(x^2)'\cos x^2.x-\sin x^2}{x^2}\\
=\dfrac{2x\cos x^2.x-\sin x^2}{x^2}\\
=\dfrac{2x^2\cos x^2-\sin x^2}{x^2}\\
b) g(x)=(x-2)\sqrt{x^2+1}\\
\Rightarrow g'(x)=\left [(x-2)\sqrt{x^2+1} \right ]'\\
=(x-2)'.\sqrt{x^2+1}+(x-2)\left (\sqrt{x^2+1} \right )'\\
=x\sqrt{x^2+1}+(x-2)\dfrac{(x^2+1)'}{2\sqrt{x^2+1}}\\
=x\sqrt{x^2+1}+(x-2)\dfrac{2x}{2\sqrt{x^2+1}}\\
=x\sqrt{x^2+1}+(x-2)\dfrac{x}{\sqrt{x^2+1}}\\
c)h(x)=(x^3-x^2+1)^{10}\\
\Rightarrow h'(x)=\left [(x^3-x^2+1)^{10} \right ]'\\
=10(x^3-x^2+1)^{9}.(x^3-x^2+1)'\\
=10(x^3-x^2+1)^{9}.(3x^2-2x)\\$
