# Math Help - Differentiate and Simplify

1. ## Differentiate and Simplify

Differentiate the following functions and simplify:

a) (x^4+2x^1/2)(lnx)

b) x^(x^2)

c) sinx/x^6

d) (e^4x^2)-2.
N:B The parentheses is not in the question, I only used it to separate -2 from x as it is not part of the power of x and to make the question clearer

2) Determine the derivative of the inverse f^-1 of f(x) =3^(x/2)+logx(base 2)+(x^3) -7 at the point 5=f(2).
Write down the equation of the tangent line of f^-1 at 5.

Happy New Year!!!!!!!!!

2. ## re: Differentiate and Simplify

you should know the chain rule and how to derive a fraction of functions or multiples of functions
$y = f(x)g(x)$

$y'= f'(x)g(x) + g'(x)f(x)$
if
$y = \frac{f(x)}{g(x)}$

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

the second one you need to make a trick let

$u = x^{x^2}$
take the ln
$\ln u = \ln x^{x^2}$

$\ln u = x^2 \ln x$ derive both sides

show some work

3. ## re: Differentiate and Simplify

2) You have to use:
$\frac{df^{-1}}{dx}(f(x))=\frac{1}{f'(x)}$

4. ## re: Differentiate and Simplify

I need you to put me through.
in number 2, the log is to base 2.
how do i even start to solve that?
Thanks

5. ## re: Differentiate and Simplify

Originally Posted by Tizi
I need you to put me through.
in number 2, the log is to base 2.
how do i even start to solve that?
Thanks
Use the change of base rule: $\log_2(x) = \dfrac{\ln(x)}{\ln(2)}$

6. ## re: Differentiate and Simplify

in a) i already solved with chain rule
here is what i did.
(x^4+2x^1/2)(lnx)
4x^3+x^(-1/2)*lnx + (x^4+2x^1/2)*1/x
how can i simplify this?

b) here's what i did as you have said
y=x^(x^2)
lny=ln x^(x^2)
1/y=x^2lnx
using chain rule,
1/y= x^2*1/x+2x*lnx
1/y=x+2xlnx
1/y=x(1+2lnx)
it doesn't seem right to me, i think i have missed something.
i need help!

7. ## Re: Differentiate and Simplify

Originally Posted by Tizi
I need you to put me through.
in number 2, the log is to base 2.
how do i even start to solve that?
Thanks
You want to find the derivative of $f^{-1}$ at the point $f(2)=5$ so you have to find:
$\frac{df^{-1}}{dx}(5)=\frac{1}{f'(2)}$

That means you have to calculate $f'(x)$ ...

8. ## Re: Differentiate and Simplify

Originally Posted by Tizi
in a) i already solved with chain rule
here is what i did.
(x^4+2x^1/2)(lnx)
4x^3+x^(-1/2)*lnx + (x^4+2x^1/2)*1/x
how can i simplify this?

b) here's what i did as you have said
y=x^(x^2)
lny=ln x^(x^2)
1/y=x^2lnx
using chain rule,
1/y= x^2*1/x+2x*lnx
1/y=x+2xlnx
1/y=x(1+2lnx)
it doesn't seem right to me, i think i have missed something.
i need help!
for the first one the second term $\frac{x^4+2 x^{\frac{1}{2}}}{x} =x^3 + 2 x^{\frac{-1}{2}}$

for the second one you have to becareful
$\ln y = x^2 \ln x$

$\frac{y'}{y} = x + 2x \ln x$
y = x^2 ln x

$y' = y ( x + 2x \ln x ) = (x^2 \ln x )(x + 2x \ln x )$

9. ## Re: Differentiate and Simplify

Originally Posted by Siron
2) You have to use:
$\frac{df^{-1}}{dx}(f(x))=\frac{1}{f'(x)}$
I would like to see a proof for it, Thanks

10. ## Re: Differentiate and Simplify

Originally Posted by Amer
I would like to see a proof for it, Thanks
Take a look here:
Inverse function theorem - Wikipedia, the free encyclopedia