MA005: Calculus I

 Once we know the derivative of $\mathrm{e}^{\mathrm{X}}$ and the Chain Rule, it is relatively easy to determine the derivative of $a^{x}$ for any $a > 0$.

$D\left(a^{x}\right)=a^{x} \cdot \ln a \quad$ for $a > 0$.

Proof: If $a > 0$, then $a^{x} > 0$ and $a^{x}=e^{\ln \left(a^{x}\right)}=e^{x \cdot \ln a}$.

$D\left(a^{x}\right)=D\left(e^{\ln \left(a^{x}\right)}\right)=D\left(e^{x \cdot \ln a}\right)=e^{x \cdot \ln a} \cdot D(x \cdot \ln a)=a^{x} \cdot \ln \mathbf{a}$.

Example 3: Calculate $\mathbf{D}\left(7^{\mathrm{X}}\right)$ and $\frac{\mathbf{d}}{\mathbf{d t}}\left(2^{\sin (\mathrm{t})}\right)$

Solution: (a) $\mathrm{D}\left(7^{\mathrm{X}}\right)=7^{\mathrm{x}} \ln 7 \approx(1.95) 7^{\mathrm{x}}$.

(b) We can write $\mathrm{y}=2^{\sin (\mathrm{t})}$ as $\mathrm{y}=2^{\mathrm{u}}$ with $\mathrm{u}=\sin (\mathrm{t})$. Using the Chain Rule,

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dt}}=2^{\mathrm{u}} \cdot \ln (2) \cdot \cos (\mathrm{t})=2^{\sin (\mathrm{t})} \cdot \ln (2) \cdot \cos (\mathrm{t})$.

Practice 3: $\quad$ Calculate $\mathbf{D}\left(\sin \left(2^{\mathrm{X}}\right)\right)$ and $\frac{\mathrm{d}}{\mathrm{dt}}\left(3^{\left(\mathrm{t}^{2}\right)}\right)$.