lecture 13 more diff

For $f(x)=\sin (\frac{5x}{4})$, what is $f^{\prime }(\frac{11\pi }{5})$?

$f^{\prime }(x)=\cos (\frac{5x}{4})(\frac{5}{4})$
$f^{\prime }(\frac{11\pi }{5})=\cos (\frac{11\pi }{4})(\frac{5}{4})$
$=(\frac{-\sqrt{2}}{2})(\frac{5}{4})$
$=\frac{-5\sqrt{2}}{8}$

Find $f^{\prime }(x)$ for $f(x)=\tan (3x+1)$.

$f^{\prime }(x)={\sec }^2(3x+1)(3)$

Find $f^{\prime }(x)$ for $f(x)=\frac{1}{(\cos (2x)+3{)}^3}$.

$\frac{d}{dx}\left[\frac{1}{f(x)}\right]=-\frac{f^{\prime }(x)}{f(x{)}^2}$

$f^{\prime }(x)=\frac{(-3)(\cos (2x)+3{)}^2(-\sin (2x)(2))}{(\cos (2x)+3{)}^6}$

$f^{\prime }(x)=\frac{(6)(\cos (2x)+3{)}^2\sin (2x)}{(\cos (2x)+3{)}^6}$

$f^{\prime }(x)=\frac{6(\sin (2x)}{(\cos (2x)+3{)}^4}$

For $f(x)=x^2+e^x-2$, find the equation of the tangent to the graph when $x=0$.

$f^{\prime }(x)=2x+e^x$
$m_T=f^{\prime }(0)=1$

$f(0)=-1$
$y=mx+c$
$y=x-1$

Find $f^{\prime \prime }(x)$ for $f(x)=x^2\cos (2x)$.

$f^{\prime }(x)=(2x)(\cos (2x))+(x^2)(-\sin (2x)(2))$
$f^{\prime \prime }(x)=2\cos (2x)\cdot -4x\sin (2x))+(4x)(-\sin (2x)))(x^2)(-2\cos (2x))$
$f^{\prime \prime }(x)=$

Find $f^{\prime }(x)$ for $f(x)=2\tan (\tan (3x+1))$. >H<

$f^{\prime }(x)=2{\sec }^2(\tan (3x+1)({\sec }^2(3x+1)(3)))$

$f^{\prime }(x)=\frac{2}{{\cos }^2(\frac{3\tan (3x+1)}{{\cos }^2(3x+1)})}$

Consider the function $f(x)=\sin (2x)$. If P is the point (x,f(x)) and Q is the point (x+h,f(x+h)), $h\ne 0$ . What is gradient of the secant PQ in terms of x and h?

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$.

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{\sin (2x+2h)-\sin (2x)}{h}$.

For $f(x)=e^x(p{x}^2+qx+r)$, the curve y=f(x) is such that the tangents at x=0 and x=1 are parallel to the x-axis. The point with coordinates (0,1) is on the curve. Find p,q and r.

$f^{\prime }(0)=0$
$f^{\prime }(1)=0$

$f(0)=1$

$f^{\prime }(x)=(e^x)(p{x}^2+qx+r)+(e^x)(2px+q)$
$f^{\prime }(0)=r+q=0$
$f^{\prime }(1)=3pe+2qe+re=0$
$f(0)=r=1$
$q=-1$
$3pe-2e+e=0$
$3p-1=0$
$p=\frac{1}{3}$

For $f(x)=x\sin (x)$, find the tangent to the graph at $x=\frac{-5\pi }{2}$.

$f^{\prime }(x)=\sin (x)+x\cos (x)$
$m_T=f^{\prime }(\frac{-5\pi }{2})=-1$
$y-y_1=m(x-x_1)$

$f(\frac{-5\pi }{2})=\frac{5\pi }{2}$
$y-\frac{5\pi }{2}=-1(x+\frac{5\pi }{2})$
$y=-x$

For $f(x)=x\sin (x)$, find $f^{\prime }(\frac{11\pi }{4})$.

For $f(x)=(\tan (x)+2{)}^2$, find $f^{\prime }(x)$.

For $f(x)=a\sqrt{3}\sin (x)+b\cos (x)$ where $a,b<0$, state the relationship between a and b if there is a local minimum at $x=\frac{\pi }{3}$.

$f^{\prime }(x)=a\sqrt{3}\cos (x)-b\sin (x)$
$f^{\prime }(\frac{\pi }{3})=0$
$0=a\sqrt{3}\cos (\frac{\pi }{3})-b\sin (\frac{\pi }{3})$
$0=\frac{\sqrt{3}}{2}a-\frac{\sqrt{3}}{2}b$
$a=b$

For $f(x)=\cos (2x)+\sin (x)+x^2$, find $f^{\prime \prime }(x)$.

$f^{\prime }(x)=-2\sin (2x)+\cos (x)+2x$
$f^{\prime \prime }(x)=-4\cos (x)-\sin (x)+2$

The average rate of change of $f(x)={\log }_2(\frac{x^3}{4})$ over the interval $2\le x\le 4$ is:

rise over run

$\frac{f(4)-f(2)}{2}=\frac{4-1}{2}=\frac{3}{2}$

The value of the gradient of curve $f(x)=\cos (x)\sin (x)$ at $x=\frac{\pi }{3}$ is:

$f^{\prime }(x)=-\sin (x)\sin (x)+\cos (x)\cos (x)$
$f^{\prime }(x)=-{\sin }^2(x)+{\cos }^2(x)$
$f^{\prime }(\frac{\pi }{3})=\frac{-3}{4}+\frac{1}{4}=\frac{-1}{2}$

For $f:[1,3]\to \mathbb{R},f(x)=x^2+kx+4$, find the value(s) of k such that the absolute maximum of the function occurs when x=1.

There can be many absolute maximums.

We know that $f(1)\ge (3)$.

$k+5\ge 3k+13$
$2k\le -8$
$k\le -4$

Let f be a one-to-one differentiable function such that f(3)=7,f(7)=8,f′(3)=2 and f′(7)=3. The function g is differentiable and g(x)=f−1(x) for all x. Find g′(7).

We know that $g(f(x))=x$
Take derivative of both sides
$g^{\prime }(f(x))f^{\prime }(x)=1$
Rearrange
$f^{\prime }(x)=\frac{1}{g^{\prime }(f(x))}$

In this case,

$g^{\prime }(7)=\frac{1}{f^{\prime }(g(7))}=\frac{1}{f^{\prime }(3)}=\frac{1}{2}$