Answer to Question #231601 in Calculus for Mamaj

Question

1. Using the chain rule differentiate the following function.

a. y=(2x2+4x-2)2

b. y=(x2+2x-4)2

2. Using the product rule differentiate the following functions.

a. y=2x2-x2(1/2x2+5)2

b. y=(x+2)2(x+3)2

3. Using the quotient rule differentiate the following functions.

a. y=x2/(3x+5)2

b. y=e2x/x2+2

4. Differentiate the following functions.

a. y=x2e2x

b. y=ln(x+2x+1)

5. Differentiate the following functions (with respect of to x).

a. fx=2x2+3x(4x2-2)

b. fx=ln2x2+3e2+x

6. Differentiate the following functions and find the value of x.

a. y=x2+3x(x-2)

b. y=x2-3/x+2

Thank You

Answer 1

1.

a.

$y'=((2x^2+4x^{-2})^2)'$

$=2(2x^2+4x^{-2})(2x^2+4x^{-2})'$

$=4(x^2+2x^{-2})(4x-8x^{-3})$

$=\dfrac{16(x^4+2)(x^4-2)}{x^5}=\dfrac{16(x^4-4)}{x^5}$

b.

$y'=((x^2+2x-4)^2)'$

$=2(x^2+2x-4)(x^2+2x-4)'$

$=2(x^2+2x-4)(2x+2)$

$=4(x+1)(x^2+2x-4)$

2.

a.

$y'=(2x^2-x^2(\dfrac{1}{2}x^2+5)^2)'$

$=4x-2x(\dfrac{1}{2}x^2+5)^2-2x^2 (\dfrac{1}{2}x^2+5)(\dfrac{1}{2}(2x)+5)$

$=4x-2x(\dfrac{1}{2}x^2+5)^2-x^3 (x^2+10)$

$=4x-\dfrac{1}{2}x^5-10x^3-50x-x^5-10x^3$

$=-\dfrac{3x^5+40x^3+92x}{2}$

b.

$y'=((x+2)^2(x+3)^2)'$

$=2(x+2)(x+3)^2+2(x+2)^2(x+3)$

$=2(x+2)(x+3)(x+3+x+2)$

$=2(x+2)(x+3)(2x+5)$

3.

a.

$y'=(\dfrac{x^2}{(3x+5)^2})'=\dfrac{2x(3x+5)^2-2x^2(3)(3x+5)}{(3x+5)^4}$

$=\dfrac{2x(3x+5-3x)}{(3x+5)^3}=\dfrac{10x}{(3x+5)^3}$

b.

$y'=(\dfrac{e^{2x}}{x^2+2})'=\dfrac{2e^{2x}(x^2+2)-2xe^{2x}}{(x^2+2)^2}$

$=\dfrac{2e^{2x}(x^2-x+2)}{(x^2+2)^2}$

4.

a.

$y'=(x^2e^{2x})'=2xe^{2x}+2x^2e^{2x}=2x(x+1)e^{2x}$

b.

$y'=(\ln(x+2x+1))'=\dfrac{3}{3x+1}$

$y'=(\ln((x+2)(x+1))'=\dfrac{x+1+x+2}{(x+2)(x+1)}$

$=\dfrac{2x+3}{(x+2)(x+1)}$

5.

a.

$f'(x)=(2x^2+3x(4x^2-2))'$

$=(12x^3+2x^2-6x)'=36x^2+4x-6$

b.

$f'(x)=(\ln^2(x^2)+3e^{2+x})'$

$=2\ln(x^2)(\dfrac{1}{x^2})(2x)+3e^{2+x}$

$=\dfrac{8\ln|x|}{x}+3e^{2+x}$

6.

a.

$y'=(x^2+3x(x-2))'=2x+3(x-2)+3x$

$=8x-6, x\in \R$

b.

$y'=(x^2-\dfrac{3}{x+2})'=2x-\dfrac{3}{(x+2)^2}, x\not=-2$