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Answer:

See below.

Explanation:

This differential equation is linear so it can be represented as the sum

#y = y_h+y_p# with

#y'_h+2/x y_h = 0#
#y'_p+2/xy_p = 2 e^(x^2)( (x^2 + 1)/x)#

The solution for #y_h# is easily determined as

#y_h = C_0/x^2#

now supposing that #y_p = (C(x))/x^2# and substituting into

#y'_p+2/xy_p = 2 e^(x^2)( (x^2 + 1)/x)#

we get

#(2 e^(x^2) (1 + x^2))/x - (C'(x))/x^2=0# or

#C'(x) = e^(x^2) x (1 + x^2)# or

#C(x) = e^(x^2)x^2+C_1# and finally

#y = ( e^(x^2)x^2+C_1)/x^2 = C_1/x^2+e^(x^2)#

Answer:

#f'(x) = 1/x - (2x)/(1+x^2) #

Explanation:

We must use our log laws:

#log_alpha(beta/gamma) -= log_alpha beta - log_alpha gamma#

#=> ln | x / (1+x^2) | = ln|x| - ln|1+x^2 | #

Now applying our knowledge of differentiating logs:

#d/(dx) ln(f(x)) = (f'(x))/f(x) #

By using the chain rule...

#d/(dx)( ln|x| - ln|1+x^2| ) = (d/(dx)(x))/x - (d/(dx)(1+x^2)) /( 1+x^2)#

#color(blue)(= 1/x - (2x)/(1+x^2) #

Answer:

#int_0^1 9/(3+x^2)^2\ dx=((2pi+3)sqrt(3))/24#

Explanation:

#\ \ \ \ \ \ int_0^1 9/(3+x^2)^2\ dx#

Substitute #x=sqrt(3)tan(theta)# and #dx=sqrt(3)sec^2(theta)d\theta#:
#=int_arctan(0)^arctan(1/sqrt(3))9/(3+(sqrt(3)tan(theta))^2)^2sqrt(3)sec^2(theta)\ d\theta#

#=sqrt(3)int_0^(pi/6) (9sec^2(theta))/(3+3tan^2(theta))^2\ d\theta#

#=sqrt(3)int_0^(pi/6) sec^2(theta)/(1+tan^2(theta))^2\ d\theta#

Using the fact that #tan^2(theta)+1=sec^2(theta)#:

#=sqrt(3)int_0^(pi/6) sec^2(theta)/sec^4(theta)\ d\theta#

#=sqrt(3)int_0^(pi/6) 1/sec^2(theta)\ d\theta#

Since #sec(theta)=1/cos(theta)#, we have

#=sqrt(3)int_0^(pi/6) cos^2(theta)\ d\theta#

Using the fact that #cos^2(theta)=(1+cos(2theta))/2#:

#=sqrt(3)/2int_0^(pi/6) 1+cos(2theta)\ d\theta#

#=sqrt(3)/2[theta+sin(2theta)/2]_0^(pi/6)#

#=sqrt(3)/2(pi/6+1/4)#

#=((2pi+3)sqrt(3))/24#

Answer:

The coordinates of the midpoint of #QR# is #=(17/2,17/4)#. The equation of the curve is #y=-4(6-2x))^(1/2)+16#

Explanation:

The gradient to the curve #f(x)# is

#dy/dx=f'(x)=4/(sqrt(6-2x))#

At the point #P(1,8)#, the gradient is

#f'(1)=4/sqrt(6-2*1)=4/sqrt4=2#

The slope of the tangent at the point #P# is #m=2#

Therefore,

The slope of the normal at the point #P#

#=m'=-1/m=-1/2#

The equation of the normal at the point #P# is

#y-8=-1/2(x-1)#

#2y-16=-x+1#

#2y+x=17#

When #x=0#, #=>#, #y=17/2#

The point #Q=(0,17/2)#

When #y=0#, #=>#, #x=17#

The point #R=(17,0)#

The midpoint of #QR# is

#=((17+0)/2,(17/2+0)/2)#

#=(17/2, 17/4)#

The equation of the curve is obtained by integrating the derivative, or by solving the differential equation

#y=int(4dx)/(sqrt(6-2x))#

#=4int(6-2x)^(-1/2)#

#=4*-1/2*2/1*(6-2x)^(1/2)+C#

#y=-4(6-2x))^(1/2)+C#

Plugging in the values of #P(1,8)#

#8=-4(6-2)^(1/2)+C#

#C=16#

The equation of the curve is

#y=-4(6-2x))^(1/2)+16#

graph{(2y+x-17)(y+4sqrt(6-2x)-16)=0 [-22.17, 23.44, -4.57, 18.27]}

Answer:

#lim_(x->0) (e^x+3x)^(1/x) = e^4#

Explanation:

Write the function as:

#(e^x+3x)^(1/x) = (e^(ln(e^x+3x)))^(1/x) = e^(ln(e^x+3x)/x)#

Consider now the limit:

#lim_(x->0) ln(e^x+3x)/x#

It is in the indeterminate form #0/0# so we can use l'Hospital's rule:

#lim_(x->0) ln(e^x+3x)/x = lim_(x->0) (d/dx ln(e^x+3x))/(d/dx x)#

#lim_(x->0) ln(e^x+3x)/x = lim_(x->0) (e^x+3)/(e^x+3x) = 4#

As the limit is finite and the function #e^x# is continuous for #x in RR# we have:

#lim_(x->0) e^(ln(e^x+3x)/x) = e^((lim_(x->0) ln(e^x+3x)/x)) = e^4#

Answer:

See below.

Explanation:

I hope it helps.

The partial derivative is intrinsically associated to the total variation.

Suppose we have a function #f(x,y)# and we want to know how much it varies when we introduce an increment to each variable.

Fixing ideas, making #f(x,y) = k x y# we want to know how much it is

#df(x,y) = f(x+dx,y+dy)-f(x,y)#

In our function-example we have

#f(x+dx,y+dy) = k(x+dx)(y+dy) = k x y + k x dx + k y dy + k dx dy#

and then

#df(x,y)=k x y + k x dx + k y dy + k dx dy-k x y = k x dx +k y dy + k dx dy#

Choosing #dx, dy# arbitrarily small then #dx dy approx 0# and then

#df(x,y) = k x dx +k y dy#

but generally

#df(x,y) = f(x+dx,y+dy)-f(x,y) =1/2(2 f(x+dx,y+dy)-2f(x,y)+f(x+dx,y)-f(x+dx,y)+f(x,y+dy)-f(x,y+dy)) =#

#=1/2(f(x+dx,y)-f(x,y))/dx dx +1/2 (f(x,y+dy)-f(x,y))/dy dy+#

#+ 1/2(f(x+dx,y+dy)-f(x,y+dy))/dx dx+1/2(f(x+dx,y+dy)-f(x+dx,y))/dy dy#

now making #dx, dy# arbitrarily small we have

#df(x,y) = 1/2(2f_x(x,y)dx+2f_y(x,y)dy) = f_x(x,y)dx+f_y(x,y) dy#

so we can compute the total variation for a given function, by calculating the partial derivatives #f_(x_1),f_(x_2), cdots, f_(x_n)# and compounding

#df(x_1,x_2,cdots,x_n) = f_(x_1)dx_1+ cdots+ f_(x_n)dx_n#

Here, the quantities #f_(x_i) # are called partial derivatives and can also be represented as

#(partial f)/(partial x_i)#

In our example

#f_x = (partial f)/(partial x) = k x# and

#f_y = (partial f)/(partial y) = k y#

NOTE

#f_x(x,y) = lim_((dx->0),(dy->0))(f(x+dx,y)-f(x,y))/dx = lim_((dx->0),(dy->0))(f(x+dx,y+dy)-f(x,y))/dx#

#f_y(x,y) = lim_((dx->0),(dy->0))(f(x,y+dy)-f(x,y))/dy = lim_((dx->0),(dy->0))(f(x+dx,y+dy)-f(x,y))/dy #

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