| 1) \(y=20x^3-12x+\dfrac 8 {10} \) | 2) \(y=9^x + \log_2 x+7\) |
| 3) \(y=\sqrt[9]x + 10x-9\) | 4) \(y=e^x+1\) |
| 5) \(y=4x \cos x\) | 6) \(y=\dfrac {10x}{\sin x}\) |
| 7) \(y=\dfrac{x^2}{2}+\dfrac{x^10}{2}\) | 8) \(y=-6x \ln x\) |
| 9) \(y= \dfrac {\ln x} x \) | 10) \(y=\dfrac 1 {2x+2}\) |
Solucions
Solucions
1
\[ y'=60x^2-12\]
2
\[ y'=9^x \cdot \ln 9 + \frac 1 {x \ln 2}\]
3
\[ y'=\dfrac 1 {9 \sqrt[9]{x^8}}+10\]
4
\[ y'=e^x\]
5
\[ y'=D(4x) \cdot \cos x + D(\cos x) \cdot 4x\]
\[ y'=4 \cdot \cos x - \sin x \cdot 4x\]
6
\[y'= \frac {D(10x) \cdot \sin x - D(\sin x) \cdot 10x }{(\sin x)^2}\]
\[y'= \frac {10 \cdot \sin x - \cos x \cdot 10x }{\sin^2 x}\]
7
\[ y'=\frac {D(x^2)}{2} +0\] No cal derivar-lo com una divisió ja que al denominador no hi ha x.
\[ y'=\frac {\not 2x}{\not 2} \]
\[ y'=x \]
8
\[ y'=D(-6x) \cdot \ln x + D(\ln x) \cdot (-6x) \]
\( y'= -6 \cdot \ln x + \dfrac 1 x \cdot (-6x) \)
\[ y'= -6 \ln x + \frac {-6x} x \]
\[ y'= -6 \ln x -6\]
9
\[ y' = \frac {D(\ln x) \cdot x - D(x) \cdot \ln x}{x^2} \]
\[ y' = \frac {\dfrac 1 x \cdot x - 1 \cdot \ln x}{x^2} \]
\[ y' = \frac {\dfrac x x - \ln x}{x^2} \]
\[ y' = \frac {1 - \ln x}{x^2} \]
10
\[y'=\frac {D(1) \cdot (2x+2) - D(2x+2) \cdot 1}{(2x+2)^2}\]
\[y'=\frac {0 \cdot (2x+2) - 2 \cdot 1}{(2x+2)^2}\]
\[y'=\frac {- 2 }{(2x+2)^2}\]