lecture 12 differentiation of transcendental functions ♡

Find derivative of $f (x) = ln (2 x - 7)$ .

$f^{'} (x) = \frac{1}{2 x - 7} \cdot (2) = \frac{2}{2 x - 7}$

For $f (x) = x^{2} - 2$ , use Newton’s method to approximate $x_{1}$ starting with $x_{0} = - 1$ .

$x_{1} = x_{0} - \frac{f ( x _{0} )}{f ^{'} ( x _{0} )}$

$x_{1} = - 1 - \frac{( - 1 ) ^{2} - 2}{2 ( - 1 )}$
$x_{1} = \frac{- 3}{2}$

Find $f^{''} (- 2)$ for $f (x) = e^{3 x} + 5 x^{2} - \frac{3}{x}$ .

$f^{'} (x) = 3 e^{3 x} + 10 x + (- 3) (- 1) (x)^{- 2}$
$f^{'} (x) = 3 e^{3 x} + 10 x + 3 x^{- 2}$
$f^{''} (x) = 9 e^{3 x} + 10 - 6 x^{- 3}$

$f^{''} (2) = 9 e^{6} + 10 + \frac{- 6}{( - 2 ) ^{3}}$
$f^{''} (2) = 9 e^{6} + \frac{40}{4} - \frac{3}{4}$
$f^{''} (2) = 9 e^{6} + \frac{37}{4}$

For $f (x) = - x$ , find angle $θ$ from the positive x axis to the tangent of f when $x = \frac{- 3}{4}$ .

$f^{'} (x) = \frac{1}{2} (- x)^{\frac{- 1}{2}} (- 1)$
$f^{'} (x) = \frac{1}{- 2 - x}$

$tan (θ) = f^{'} (\frac{- 3}{4}) = \frac{1}{- 2 ( \frac{3}{2} )}$
$tan (θ) = \frac{1}{\frac{- 3}{1}} = \frac{- 1}{3}$

$θ = \frac{5 π}{6}$

Let $f (x) = - ln (x + 3)$ . A tangent to the graph has y-intercept at $(0, c)$ . What is the maximum value of c?

$f^{'} (x) = \frac{- 1}{x + 3}$

Find the coordinate of the point of inflection of $f (x) = - \frac{x ^{3}}{2} - \frac{x ^{2}}{4} - x$ .

$f^{'} (x) = \frac{- 3 x ^{2}}{2} + \frac{- x}{2} - 1$
$f^{''} (x) = - 3 x - \frac{1}{2} = 0$
$x = \frac{- 1}{6}$ is a possible POI. Testing concavity before and after.
$f^{''} (0) < 0$
$f^{''} (\frac{- 1}{3}) > 0$

Concavity changes, hence is POI at $x = \frac{- 1}{6}$

$f (\frac{- 1}{6}) = \frac{35}{216}$

Find the equation of the tangent to the curve of $f (x) = (2 x + 3)^{\frac{3}{5}} + 3$ at $x = \frac{- 3}{2}$ .

$y - y_{1} = m (x - x_{1})$

$m = f^{'} (\frac{- 3}{2})$

$f^{'} (x) = \frac{3}{5} (2 x + 3)^{\frac{- 2}{5}} (2)$
$f^{'} (x) = \frac{6}{5} (2 x + 3)^{\frac{- 2}{5}}$
$f^{'} (\frac{- 3}{2}) = 0$

A point is $(\frac{- 3}{2}, 3)$

$y - 3 = 0$
$y = 3$

Differentiate $f (x) = \frac{4 e ^{2 x} - 2 e ^{x} + 1}{2 e ^{2 x}}$ .

$f^{'} (x) = \frac{( 8 e ^{2 x} ) ( - 2 e ^{x} ) ( 2 e ^{2 x} ) - ( 4 e ^{2 x} - 2 e ^{x} + 1 ) ( 4 e ^{2 x} )}{4 e ^{4 x}}$

Differentiate $f (x) = e^{2 x} + e^{4} + e^{- 2 x}$ .

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

Differentiate $f (x) = 3 ln (4 x) + x$ .

$f^{'} (x) = 3 (\frac{4}{4 x}) + 1$
$f^{'} (x) = \frac{3}{x} + 1$

Find $f^{''} (x)$ for $f (x) = 36 x e^{- 0.5 x}$ .

$f (x) = (36 x) (e^{- 0.5 x})$
$f^{'} (x) = (36) (e^{- 0.5 x}) + (36 x) (- 0.5 e^{- 0.5 x})$
$f^{'} (x) = 36 e^{- 0.5 x} - 18 x e^{- 0.5 x}$
$f^{''} (x) = - 18 e^{- 0.5 x} + (- 18) (e^{- 0.5 x}) + (- 18 x) (- 0.5 e^{- 0.5 x})$
$f^{''} (x) = (9 x - 36) e^{- 0.5 x}$

What is the equation of the tangent to the graph $y = e^{x} - 2 x$ at the point $(0, 1)$ ?

$\frac{d y}{d x} = \frac{1}{2} (e^{x})^{\frac{- 1}{2}} (e^{x}) - 2$
$\frac{d y}{d x} = \frac{1}{2 e ^{x}} (e^{x}) - 2$

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lecture 12 differentiation of transcendental functions

Find derivative of $f (x) = ln (2 x - 7)$ .

For $f (x) = x^{2} - 2$ , use Newton’s method to approximate $x_{1}$ starting with $x_{0} = - 1$ .

Find $f^{''} (- 2)$ for $f (x) = e^{3 x} + 5 x^{2} - \frac{3}{x}$ .

For $f (x) = - x$ , find angle $θ$ from the positive x axis to the tangent of f when $x = \frac{- 3}{4}$ .

Let $f (x) = - ln (x + 3)$ . A tangent to the graph has y-intercept at $(0, c)$ . What is the maximum value of c?

Find the coordinate of the point of inflection of $f (x) = - \frac{x ^{3}}{2} - \frac{x ^{2}}{4} - x$ .

Find the equation of the tangent to the curve of $f (x) = (2 x + 3)^{\frac{3}{5}} + 3$ at $x = \frac{- 3}{2}$ .

Differentiate $f (x) = \frac{4 e ^{2 x} - 2 e ^{x} + 1}{2 e ^{2 x}}$ .

Differentiate $f (x) = e^{2 x} + e^{4} + e^{- 2 x}$ .

Differentiate $f (x) = 3 ln (4 x) + x$ .

Find $f^{''} (x)$ for $f (x) = 36 x e^{- 0.5 x}$ .

What is the equation of the tangent to the graph $y = e^{x} - 2 x$ at the point $(0, 1)$ ?

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wiki ♡

Explorer

lecture 12 differentiation of transcendental functions

Find derivative of f(x)=ln(2x−7).

For f(x)=x2−2, use Newton’s method to approximate x1​ starting with x0​=−1.

Find f′′(−2) for f(x)=e3x+5x2−x3​.

For f(x)=−x​, find angle θ from the positive x axis to the tangent of f when x=4−3​.

Let f(x)=−ln(x+3). A tangent to the graph has y-intercept at (0,c). What is the maximum value of c?

Find the coordinate of the point of inflection of f(x)=−2x3​−4x2​−x.

Find the equation of the tangent to the curve of f(x)=(2x+3)53​+3 at x=2−3​.

Differentiate f(x)=2e2x4e2x−2ex+1​.

Differentiate f(x)=e2x+e4+e−2x.

Differentiate f(x)=3ln(4x)+x.

Find f′′(x) for f(x)=36xe−0.5x.

What is the equation of the tangent to the graph y=ex​−2x at the point (0,1)?

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Table of Contents

Find derivative of $f (x) = ln (2 x - 7)$ .

For $f (x) = x^{2} - 2$ , use Newton’s method to approximate $x_{1}$ starting with $x_{0} = - 1$ .

Find $f^{''} (- 2)$ for $f (x) = e^{3 x} + 5 x^{2} - \frac{3}{x}$ .

For $f (x) = - x$ , find angle $θ$ from the positive x axis to the tangent of f when $x = \frac{- 3}{4}$ .

Let $f (x) = - ln (x + 3)$ . A tangent to the graph has y-intercept at $(0, c)$ . What is the maximum value of c?

Find the coordinate of the point of inflection of $f (x) = - \frac{x ^{3}}{2} - \frac{x ^{2}}{4} - x$ .

Find the equation of the tangent to the curve of $f (x) = (2 x + 3)^{\frac{3}{5}} + 3$ at $x = \frac{- 3}{2}$ .

Differentiate $f (x) = \frac{4 e ^{2 x} - 2 e ^{x} + 1}{2 e ^{2 x}}$ .

Differentiate $f (x) = e^{2 x} + e^{4} + e^{- 2 x}$ .

Differentiate $f (x) = 3 ln (4 x) + x$ .

Find $f^{''} (x)$ for $f (x) = 36 x e^{- 0.5 x}$ .

What is the equation of the tangent to the graph $y = e^{x} - 2 x$ at the point $(0, 1)$ ?