Question

differentiate each of the following functions and simplify.

1. f(x)= (1+√x) (2+√x)
2. h(x)= (5x^2-1)^8 (1-6x^2)^4
3. h(x)= (x^2(1+x^2)^1/2)^1/2)
4. y= (1 - cosx / 1 + cosx)
5. y= tan (xsinx)
6. y= tan^-1 (a+x/1-ax)
7. y= (1-x^2) cos^-1x - x
8. y= cot^-1 (x^2-1)^1/2 + sec^-1x
9.y= x/(a^2-x^2)^1/2 - sin^-1 x/a
10. f(x)= sin^2 ( cos 2x)

Answer 1

Do you know how to do any of them by chance?

Answer 2

1. First expand, then use the fact that:
\[\frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\]
2. Use the chain rule, for example:
\[\frac{d}{dx}(2x+3)^3=\frac{d}{dx}(2x+3)\cdot \frac{d}{du}(u^3)=2\cdot 3(2x+3)^2=6(2x+3)^2\]
Where u=2x+3
3. Again, use the chain rule, but multiple times. This time, however, you will also have to use the product rule:
\[\frac{d}{dx}(f(x)g(x))=f(x)\cdot\frac{d}{dx}(g(x))+g(x)\cdot\frac{d}{dx}(f(x))\]
4. You will have to use the quotient rule:
\[\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)\cdot\frac{d}{dx}(f(x))-f(x)\cdot\frac{d}{dx}(g(x))}{(g(x))^2}\]
5. First remember that tan x = (sin x)/(cos x), then use product rule on x sin x, then chain rule on tan(x sin x), then quotient rule on sin(x sin x) / cos(x sin x)
6. First, note:
\[\frac{d}{dx}(\tan^{-1} x)=\frac{1}{x^2 +1}\]
Then use quotient rule and chain rule
7. First note:
\[\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))\]
\[\frac{d}{dx}(\cot^{-1} x)=-\frac{1}{\sqrt{1-x^2}}\]
And then use product rule
8. Note:
\[\frac{d}{dx}(\cot^{-1} x)=-\frac{1}{x^2 +1}\]
\[\frac{d}{dx}(\sec^{-1} x)=\frac{1}{x^2 \sqrt{1-\frac{1}{x^2}}}\]
9. First note:
\[\frac{d}{dx}(\sin^{-1} x)=\frac{1}{\sqrt{1-x^2}}\]
Then use simple chain rule for the first half
10. Use chain rule (twice), and note:
\[\sin^2 x=\sin{x}\sin{x}\]
So then use the product rule