Can someone help me revise how to do these...
x^4cos(x^2)
tan(x)/1+x^3 - quotient rule
sin(exp(1+x^2)) - chain rule?
1) $\displaystyle x^4\cos x^2$
Product Rule:$\displaystyle (x^4\cos x^2)' = (x^4)'\cos x^2 + x^4(\cos x^2)'$
Apply chain rule on $\displaystyle \cos x^2$
$\displaystyle = 4x^3\cos x^2+ x^4 (2x\cos x^2)$
$\displaystyle = 4x^3\cos x^2 + 2x^5\cos x^2$
2)$\displaystyle \frac{\tan x}{1+x^3}$
Yes try quotient rule, and tell us what you get.
3) $\displaystyle z = \sin e^{1+x^2}$
Yes chain rule
First sub $\displaystyle u = 1+x^2, y = e^u$, then $\displaystyle z = \sin e^{1+x^2} = \sin y$
$\displaystyle \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{du}\frac{du}{dx}$
So $\displaystyle \frac{dz}{dy} = \cos y, \frac{dy}{du} = e^u, \frac{du}{dx} = 2x$,
$\displaystyle \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{du}\frac{du}{dx} = (\cos y)(e^u)(2x) = 2x e^{1+x^2} \sin e^{1+x^2}$
$\displaystyle
\begin{gathered}
u = \tan (x) \hfill \\
v = 1 + x^3 \hfill \\
\hfill \\
\frac{{du}}
{{dx}} = \sec ^2 (x) \hfill \\
\frac{{dv}}
{{dx}} = 3x^2 \hfill \\
\end{gathered}
$
$\displaystyle
\begin{gathered}
so \hfill \\
\frac{{3x^2 \tan (x) - 1 + x^3 \sec ^2 (x)}}
{{(1 + x^3 )^2 }} ? \hfill \\
\hfill \\
\end{gathered}
$
No. Two problems: first, the quotient rule is $\displaystyle \left(\frac uv\right)' = \frac{vu' - uv'}{v^2}$; your numerator is backwards. Second, you did not properly distribute the $\displaystyle \sec^2x$ over the $\displaystyle 1 + x^3$ factor (or did you forget to type the parentheses?)
You should get:
$\displaystyle \frac d{dx}\left[\frac{\tan x}{1+x^3}\right]=\frac{\left(1 + x^3\right)\sec^2x - 3x^2\tan x}{\left(1 + x^3\right)^2}$
Yes, it should be $\displaystyle 2xe^{1+x^2}\cos (e^{1+x^2})$. See below:
__________________
$\displaystyle z=\sin e^{1+x^2}$
$\displaystyle \frac{\mathrm{d}z}{\mathrm{d}x} = \frac{\mathrm{d}z}{\mathrm{d}u}.\frac{\mathrm{d}u} {\mathrm{d}x}$
Let $\displaystyle u=e^{1+x^2}$ $\displaystyle \therefore z=\sin u$.
$\displaystyle \frac{\mathrm{d}z}{\mathrm{d}x} = \frac{\mathrm{d}(\sin u)}{\mathrm{d}u}.\frac{\mathrm{d}(e^{1+x^2})}{\mat hrm{d}x}$
$\displaystyle =\cos u . \frac{\mathrm{d}(e^{1+x^2})}{\mathrm{d}x} = \cos (e^{1+x^2}) . \frac{\mathrm{d}(e^{1+x^2})}{\mathrm{d}x} $
For $\displaystyle \frac{\mathrm{d}(e^{1+x^2})}{\mathrm{d}x}$, use chain rule with $\displaystyle v=1+x^2$
$\displaystyle \therefore \frac{\mathrm{d}(e^{1+x^2})}{\mathrm{d}x} = 2xe^v = 2xe^{1+x^2}$
$\displaystyle \frac{\mathrm{d}z}{\mathrm{d}x} = 2xe^{1+x^2}\cos (e^{1+x^2}) $
hey ppl ..... just joined this forum a lil while ago and i'm impressed by what i've seen so far. although i havent done calculus since 5 years ago (so i'm really really rusty! ) but just looking at the recent posts brings back some wonderful memories.
amm .... i'm new to all of this but can someone tell me how to get the different mathematical symbols when posting? i mean like how to get the dy/dx or integral sign, exponent, fractions etc. is there a clipboard on the forum with the symbols on it or are they from an external program or what?
thanks