Derivative – Solved Problems

Problem 2-11: Differentiating by Chain Rule

Differentiate $\left(1+\sqrt{x}\right)^{8}$ by using the chain rule.

April 11, 2010

Problem 2-10: Diffrentiating

Diffrentiate
a) $f_1\left(x\right)= \displaystyle\frac{1-2x^2}{1+x+x^2}$
b) $f_2\left(x\right)= (1-2x^2)(1+x+x^2)$

Solution
The following rules can be used:
I) $\displaystyle\frac{d}{dx}ax^n=anx^{n-1}$
II) $\displaystyle\frac{d}{dx}\left(u+v\right)=\frac{du}{dx}+\frac{dv}{dx}$

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April 11, 2010

Problem 2-9: Differentiating Polynomial and Rational Functions

Differentiate
a) $f\left(x\right)= 1+ 2x$
b) $f\left(x\right)= 1 - 2x + 2x^2$
c) $f\left(x\right)= 1- x^2 +\frac{1}{x}$
d) $f\left(x\right)= 1 - x +\frac{1}{1-x}$

Solution
The following rules can be used:
I) $\displaystyle\frac{d}{dx}ax^n=anx^{n-1}$
II) $\displaystyle\frac{d}{dx}\left(u+v\right)=\frac{du}{dx}+\frac{dv}{dx}$
III) $\displaystyle\frac{d}{dx}\frac{u}{v}=\frac{1}{v^2}\left(v\frac{du}{dx}-u\frac{dv}{dx}\right)$
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April 11, 2010

Problem 2-2: Evaluating Derivative of Functions and the Tangent Lines

Find the derivative of $f(x)$ and the equation of the tangent line at $x_0=-1$ .

a) $f(x)=x^2$
b) $f(x)=x^3+x+1$
c) $f(x)=\frac{1}{x}$

Solution
The equation of the tangent line at ${x}_{0}$ is $y = f'(x_0) (x-x_0) + f(x_0)$ .
a) $f(x)=x^2, f'(x)=2x$
$y = f'(x_0) (x-x_0) + f(x_0) = 2x_0 (x-x_0)+x_0^2 = 2x_0 x-x_0^2$
$y = -2x-1$ .

b) $f(x)=x^3+x+1$

$f'(x)=3x^2+1$
$y=f'(x_0) (x-x_0) + f(x_0) =(3x_0^2+1) (x-x_0)+x_0^3+x_0+1=(3x_0^2+1)x-2x_0^3+1$
$y=4x+3$ .

c) $f(x)=\frac{1}{x}=x^{-1}$
$f'(x)=-x^{-2}$ $f'(x)=-\frac{1}{x^2}$
$y=f'(x_0) (x-x_0) + f(x_0) =-\frac{1}{x_0^2}(x-x_0)+\frac{1}{x_0}=-\frac{1}{x_0^2} x +\frac{2}{x_0}$
$y = -x-2$

April 7, 2010

Category: Derivative

Problem 2-11: Differentiating by Chain Rule

Problem 2-10: Diffrentiating

Problem 2-9: Differentiating Polynomial and Rational Functions

Problem 2-2: Evaluating Derivative of Functions and the Tangent Lines