# Thread: Help to find solutions

1. ## Help to find solutions

1. suppose that f(x) and g(x) are twise differentiable at x=a,
show that
(fg)"(a)= f(a) g"(a) + 2f'(a) g'(a) + f"(a) g(a)
__________________________________________________ _____________

2. Use the definition of derivative to show that for the function

f(x)=3(x)^(1/3)

(f of x is equal to 3 x to the power 1/3)

we have f'(a)=a^(2/3), for every "a" belongs to "R" (Set of real numbers)

[Use the identity (a^3)-(b^3)= (a-b)((a^2)+ab+(b^2))]
__________________________________________________ _____________

3. Differentiate the following and prove the necessary expressions as given

a) If y=xtanx

show that

x sin²x (dy/dx)=x²tan²x +y sin²x
_____
b) y= (x+√ x²-1 )ⁿ
prove that

(x²-1) (dy/dx)²=n²y²

Plz help me to solve these.

2. nobody

3. Originally Posted by chanaka89
2. Use the definition of derivative to show that for the function

f(x)=3(x)^(1/3)

(f of x is equal to 3 x to the power 1/3)

we have f'(a)=a^(2/3), for every "a" belongs to "R" (Set of real numbers)

[Use the identity (a^3)-(b^3)= (a-b)((a^2)+ab+(b^2))]
_________________________________________________
$\displaystyle f(x) = 3x^{\frac{1}{3}}$.

$\displaystyle f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$

$\displaystyle = \lim_{h \to 0}\frac{3(x + h)^{\frac{1}{3}} - 3x^{\frac{1}{3}}}{h}$

$\displaystyle = \lim_{h \to 0}\frac{3[(x + h)^{\frac{1}{3}} - x^{\frac{1}{3}}]}{h}$

$\displaystyle = \lim_{h \to 0}\frac{3\left[\frac{(x + h) - x}{(x + h)^{\frac{2}{3}} + + (x + h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}\right]}{h}$

$\displaystyle = \lim_{h \to 0}\frac{\frac{3h}{(x + h)^{\frac{2}{3}} + (x + h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}}{h}$

$\displaystyle = \lim_{h \to 0}\frac{3}{(x + h)^{\frac{2}{3}} + (x + h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}$

$\displaystyle = \frac{3}{x^{\frac{2}{3}} + x^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}$

$\displaystyle = \frac{3}{3x^{\frac{2}{3}}}$

$\displaystyle = \frac{1}{x^{\frac{2}{3}}}$

$\displaystyle = x^{-\frac{2}{3}}$

4. Originally Posted by Prove It
$\displaystyle f(x) = 3x^{\frac{1}{3}}$.

$\displaystyle f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$

$\displaystyle = \lim_{h \to 0}\frac{3(x + h)^{\frac{1}{3}} - 3x^{\frac{1}{3}}}{h}$

$\displaystyle = \lim_{h \to 0}\frac{3[(x + h)^{\frac{1}{3}} - x^{\frac{1}{3}}]}{h}$

$\displaystyle = \lim_{h \to 0}\frac{3\left[\frac{(x + h) - x}{(x + h)^{\frac{2}{3}} + + (x + h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}\right]}{h}$

$\displaystyle = \lim_{h \to 0}\frac{\frac{3h}{(x + h)^{\frac{2}{3}} + (x + h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}}{h}$

$\displaystyle = \lim_{h \to 0}\frac{3}{(x + h)^{\frac{2}{3}} + (x + h)^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}$

$\displaystyle = \frac{3}{x^{\frac{2}{3}} + x^{\frac{1}{3}}x^{\frac{1}{3}} + x^{\frac{2}{3}}}$

$\displaystyle = \frac{3}{3x^{\frac{2}{3}}}$

$\displaystyle = \frac{1}{x^{\frac{2}{3}}}$

$\displaystyle = x^{-\frac{2}{3}}$
Thanks

5. Originally Posted by chanaka89
1. suppose that f(x) and g(x) are twise differentiable at x=a,
show that
(fg)"(a)= f(a) g"(a) + 2f'(a) g'(a) + f"(a) g(a)
Use the product rule- twice.

__________________________________________________ _____________

2. Use the definition of derivative to show that for the function

f(x)=3(x)^(1/3)

(f of x is equal to 3 x to the power 1/3)

we have f'(a)=a^(2/3), for every "a" belongs to "R" (Set of real numbers)

[Use the identity (a^3)-(b^3)= (a-b)((a^2)+ab+(b^2))]
__________________________________________________ _____________

3. Differentiate the following and prove the necessary expressions as given

a) If y=xtanx

show that

x sin²x (dy/dx)=x²tan²x +y sin²x
Okay, what did you get for dy/dx?
_____
b) y= (x+√ x²-1 )ⁿ
prove that

(x²-1) (dy/dx)²=n²y²
Again, what did you get for dy/dx?

Plz help me to solve these.