Derivatives

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Math › Derivatives

Questions 1 - 10

$\begin{align*}&\text{Find the 2nd derivative of the function }\&f(x)=-14e^{(16x)}\end{align*}$

$-3584e^{(16x)}$

$50176e^{(32x)}$

$-224e^{(16x)}$

$802816e^{(32x)}$

Explanation

$\text{To find the second derivative of a function, one need only be}\&\text{cautious in performing the steps of derivation twice. For the}\&\text{function we're given,}\text{we'll wish to use the following:}\&d[e^{ax}]=ae^{ax}dx\&\text{Our first derivative is thus:}\&-224e^{(16x)}\&\text{And our second is:}\&-3584e^{(16x)}$

Find the derivative $\small f'(x)$ given the function $\small f(x)=2^x$

$\small f'(x)=2^x\cdot \ln 2$

$\small \small \small f'(x)=2^x$

$\small \small f'(x)=2^x\cdot e^2$

$\small \small f'(x)=e^{2x}$

Explanation

We can find the derivative $\small f'(x)$ given the function $\small f(x)=2^x$ by rewriting $\small f$ as

$\small \small f(x)=2^x=e^{\ln 2^x}$

and using the properties of logarithms ( $\small \ln a^b=b\ln a$ in particular) to get

$\small \small \small f(x)=2^x=e^{\ln 2^x}=e^{x\ln 2}$

so now we can use the chain rule

$\small \small [g(h(x))]'=g'(h(x))h'(x)$

with $\small g(x)=e^x$ and $\small h(x)=x\cdot \ln 2$ to get

$\small f'(x)=e^{x\ln 2}\cdot \ln 2=2^x\cdot \ln 2$

Find the derivative of the following function at :

$f(x)=\frac{e^x\cos(x)}{3}$

$\frac{1}{3}$

$-\frac{1}{3}$

Explanation

The derivative of the function is

$f'(x)=\frac{1}{3}(e^x\cos(x)-e^x\sin(x))$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Evaluated at the point x=0, we get

$f'(0)=\frac{1}{3}(1-0)=\frac{1}{3}$ .

Find the first derivative of the given function

Explanation

In order to find the first derivative

we must derive both sides of the equation since

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}f(x)$

$=\frac{\mathrm{d} }{\mathrm{d} x}[sin(x)]$

From the definition of the derivative of the sine function we have

$\frac{\mathrm{d} }{\mathrm{d} x}[sin(x)]=cos(x)$

As such, we have

Find the derivative of the following function at :

$f(x)=\frac{e^x\cos(x)}{3}$

$\frac{1}{3}$

$-\frac{1}{3}$

Explanation

The derivative of the function is

$f'(x)=\frac{1}{3}(e^x\cos(x)-e^x\sin(x))$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Evaluated at the point x=0, we get

$f'(0)=\frac{1}{3}(1-0)=\frac{1}{3}$ .

Find the derivative of $\small lnx^{lnx}$ .

$\small \small lnx^{lnx}(ln(lnx)/x+1/x)$

$\small \small lnx^{lnx}ln(lnx)/x$

$\small lnxlnx/x$

$\small (lnxlnx^{lnx-1})/x$

$\small \small lnx^{lnx}(ln(lnx)+1/x)$

Explanation

To solve this derivative, we need to use logarithmic differentiation. This allows us to use the logarithm rule $\small lna^{b}=blna$ to solve an easier derivative.

Let $\small y=lnx^{lnx}$ .

Now we'll take the natural log of both sides to get

$\small lny=ln(lnx^{lnx})=lnxln(lnx)$ .

Now we can use implicit differentiation to solve for $\small y'$ .

The derivative of $\small lny$ is $\small y'/y$ , and the derivative of $\small \small \small lnx(ln(lnx))$ can be found using the product rule, which states