已知向量a=(2cosx/2,1+tan^2x),b=(根号2sin(π/4+x/2),cos^2x),令f(x)=a*b1 求f(x)在【0,π/2】上的单调区间2若f(a)=11/4,a属于(π/2,π),求F(-a)的值_百度作业帮
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已知向量a=(2cosx/2,1+tan^2x),b=(根号2sin(π/4+x/2),cos^2x),令f(x)=a*b1 求f(x)在【0,π/2】上的单调区间2若f(a)=11/4,a属于(π/2,π),求F(-a)的值
已知向量a=(2cosx/2,1+tan^2x),b=(根号2sin(π/4+x/2),cos^2x),令f(x)=a*b1 求f(x)在【0,π/2】上的单调区间2若f(a)=11/4,a属于(π/2,π),求F(-a)的值
根据向量运算法则f(x)= 2 cosx/2 * 根号2sin(π/4+x/2) + cos^2x+tan^2x * cos^2x = 2*根号2 * sin(π/4+x/2)*cosx/2 +cos^2x+ sin^2 x= 2*根号2 * 1/2 * [sin (x+π/4) + sin(π/4) ] + 1= 根号2 * sin (x+π/4) + 21.f(x)在【0,π/2】上的单增区间 为 (0,π/4),单减区间为 （ π/4,π/2）2.f(a)=11/4 ,即 根号2 * sin (a+π/4) + 2 = 11/4sin (a+π/4)= 3根号2 /8 ( 可知 a+π/4 在第二象限）cos (a+π/4)= - 根号46 /8 f(-a)= 根号2 * sin (-a+π/4) + 2 = 根号2 * cos (a+π/4) +2= 2- 根号23 /4---------------------------------------------------------------------求证：1.sin^4x+cos^4x=1-2sin^2cos^2x2.1-tan^2x/1+tan^2x=cos^2x-sin^2x3.已知tanx=-1/3,求7sinx-3cosx/4sinx+5cosx的值.第一二题是求证，第三题不是求证，是求值！_百度作业帮
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求证：1.sin^4x+cos^4x=1-2sin^2cos^2x2.1-tan^2x/1+tan^2x=cos^2x-sin^2x3.已知tanx=-1/3,求7sinx-3cosx/4sinx+5cosx的值.第一二题是求证，第三题不是求证，是求值！
求证：1.sin^4x+cos^4x=1-2sin^2cos^2x2.1-tan^2x/1+tan^2x=cos^2x-sin^2x3.已知tanx=-1/3,求7sinx-3cosx/4sinx+5cosx的值.第一二题是求证，第三题不是求证，是求值！
（1）sin^4 x+cos^4 x＝（sin^2 x+cos^2 x）^2－2sin^2 xcos^2 x =1-2sin^2 xcos^2 x (2)(1-tan^2 x)/(1+tan^2x)=(cos^2 x/cos^2 x-sin^2 x/cos^2 x)/)=(cos^2 x/cos^2 x+sin^2 x/cos^2 x)=[(cos^2 x-sin^2 x )/cos^2 x×cos^2 x/(cos^2 x)sin^2 x )=cos^2 x-sin^2 x (3)因为tanx=-1/3,所以：cosx≠0把（7sinx-3cosx）/（4sinx+5cosx）分子、分母同时除以cosx,得：（7sinx-3cosx）/（4sinx+5cosx）＝（7sinx/cosx-3cosx/cosx）/（4sinx/cosx+5cosx/cosx）=(7tanx-3)/(4tanx+5)=[7×(-1/3)-3]/[4×(-1/3)+5]=-16/11注意：tanx=sinx/cosx
3.分子分母同时除以sinx得7-3tanx/4+5tanxtanx=-1/3原式=40/17
SIN^4X+COS^4X=(sin^2x+cos^2x)^2-2sin^2xcos^2x =1-(2sinxcosx)^2/2
sin^4x+cos^4x=(sin^2x)^2 +(cos^2x)^2 +2sin^2xcos^2x -2sin^2xcos^2x=(sin^2x+cos^2x)-2sin^2xcos^2x=1-2sin^2xcos^2x1-tan^2x/1+tan^2x=(1-sin^2x/cos^2x)/(1+sin^2x/cos^2x)=(cos...已知函数f(x)=2sin(x+a/2)cos(x+a/2)+2倍根号3co s^2(x+a/2)-根号3 (1)化简f(x)的解析式 (2)若0_百度作业帮
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已知函数f(x)=2sin(x+a/2)cos(x+a/2)+2倍根号3co s^2(x+a/2)-根号3 (1)化简f(x)的解析式 (2)若0
已知函数f(x)=2sin(x+a/2)cos(x+a/2)+2倍根号3co s^2(x+a/2)-根号3 (1)化简f(x)的解析式 (2)若0
（1）化简f(x)=2sin(2x+a+π/3)（2）因为a∈【0,π】,故a+π/3=π/2,可得a=π/6（3）a=π/6时,f(x)=2cos2x.令f(x)=1,则2x=π/3+2kπ或-π/3+2kπ,取k=0、1,故x=π/6或者x=5π/6.An Approach
C A L C U L U S
THE CHAIN RULE
The derivative
a function of a function
f (x) = x5 &and & g(x) = x2+ 1.
If we now let g(x) be the
of f, then f will be a function of g.
f (g(x)) = (x2+ 1)5.
What is the derivative of
&f (g(x)) ?
First, note that
&d f(x)&&&dx
That is: &The derivative of f with respect to its argument (which in this case is x) is equal to 5 times the 4th power of the argument.
This means that if g -- or any variable -- is the argument of &f, the same form applies:
&d f(g)&&&dg
&d f(h)&&&dh
&d f(v)&&&dv
In other words, we can really take the derivative of a function of an argument &only with respect to that argument.
Therefore, since &g = x2+ 1,
&d f(g)&&&dg
&= &5(x2+ 1)4.
is 2x. &What is called the chain rule states the following:
df(g(x))&&& dx
df(g)&&&dg
"If &f is a function of g& and g is a function of x,
then the derivative of &f with respect to xis equal to the derivative of &f(g) with respect to gtimes the derivative of g(x) with respect to x."
according to the chain rule, the derivative of
5(x2+ 1)4&&2x.
Note: &In &(x2+ 1)5, & x2+ 1 &is "inside" the 5th power, which is "outside." &We take the derivative from outside to inside. &When we take the outside derivative, we do not change what is inside. &We then multiply by the derivative of what is inside.
To decide which function is outside, decide which you would have to evaluate last.
To evaluate
you would first have to evaluate x2+ 1. &Then you would take its 5th power. &The 5th power therefore is outside. That is why we take that derivative first.
When we write f(g(x)), &f is outside g. We take the derivative of f with respect to g first.
Example 1.&& &f(x)
. &What is its derivative?
Solution.&&&This has the form f (g(x)). &What function is f, that is, what is outside, and what is g, which is inside?
g is x4 & 2 because that is inside the square root function, which is f. &The derivative of the square root is given in the
of Lesson 6. &For any argument g of the square root function,
Here, &g is x4 & 2. & Therefore, since the derivative of &x4 & 2 &is 4x3,
&=&½(x4 & 2)&½&&4x3 = 2x3(x4 & 2)&½.
Example 2.&&&What is the derivative of &y = sin3x ?
Solution.&&&This is the 3rd power of sin x.
&To decide which function is outside, how would you evaluate that?
You would first evaluate sin&x, and then take its 3rd power. &sin&x is inside the 3rd power, which is outside.
Now, the derivative of the 3rd power -- of g3 -- is 3g2. &Therefore, accepting for the moment that the derivative of &sin x& is&cos&x& (), the derivative of sin3x -- from outside to inside -- is
3 sin2x&&cos x.
Example 3.&&&What is the derivative of &
&& &1& &&x3 + 1
&&Solution.&&&x3 + 1 &is inside the function &
&=&x&1, whose derivative
is &x&2&;& (, Lesson 4). &We have, then,
&& &1& &&x3 + 1
(x3 + 1)&1
Therefore, its derivative is
&(x3 + 1)&2&&3x2
Example 4.&&&
that y is a function of x. & y = y(x).& Apply the
&&Solution.& &&
y, which we are assuming to be a function of x, is inside the function y2. &The derivative of y2with respect to y is 2y. &As for the derivative of
&&y with respect to x, we can only indicate it as&
. &(See .)
Problem 1.&&&Calculate the derivative of &(x2&3x + 5)9.
To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload").Do the problem yourself first!
9(x2&3x + 5)8(2x & 3)
Problem 2.&&&Calculate the derivative of (x4 & 3x2+ 4)2/3.
2/3(x4 & 3x2+ 4)&1/3(4x3 & 6x)
Problem 3.&&&Calculate the derivative of sin5x.
5 sin4x cos x
Problem 4.&&&Calculate the derivative of sin x5.
The inside function is x5 -- you would evaluate that last. &The outside function is sin x. &(This is the sine of x5.) &Therefore, the derivative is
cos x5&&5x4.
Problem 5.&&&Calculate the derivative of &sin (1 + 2).
cos (1 + 2)x&1/2.
Problem 6.&&&Calculate the derivative of &
¼(sin x)&3/4 cos x.
Example 5. &More than two functions.&&&The chain rule can be extended to more than two functions. &For example, let
The outside function is the square root. &Inside that is (1&+&a&2nd&power).& And inside that is sin x.
The derivative therefore is
½(1 + sin2x)&1/2&&2 sin x&&cos x
sin x cos x
Problem 7.&&&Calculate the derivative of &
(Compare .)
&[sin (x2+ 5)]&2&&cos (x2+ 5)&&2x
2x cos (x2+ 5)sin2(x2+ 5)&&&
Problem 8.&&&Calculate the derivative of &
Problem 9.&&&Assume that y is a function of x, and apply the chain rule to express each derivative with respect to x.
½y&½&
the chain rule
To prove the chain rule let us go back to basics. &Let f &be a
of g, which in turn is a function of x, so that we have &f(g(x)). &Then when the value of g changes by an amount &Dg, &the value of f will change by an amount &Df. &We will have the ratio
Again, since g is a function of x, then when x changes by an amount &Dx, &g will change by an amount &Dg. &We will have the ratio
But the change in x affects f &because it depends on g. &We will have
. &It will be the product of those ratios:
Let us now take the limit as &Dx approaches 0. &Then the change in g(x) -- &Dg --
will also approach 0. &Therefore, since the limit of a product is equal to the product of the limits (), and by definition of the :
This is the chain rule.
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