A few calculus derivative problems?

Question

Find the derivative. Solve all or any you might know. Please show work

a) y= 10^(Tan Pi Theta)

b) y= sqrt[t ln (t^4)]

c)y=arctan(arcsin x^.5)

d)xe^y=y-1

Answer 1

a) y= 10^(tan (πθ))
= e^(ln10*tan(πθ))
= e^u where u = ln10*tan(πθ
dy/dθ =dy/du * du/dθ
= e^u * ln10*πsec²(πθ)
= ln10*πsec²(πθ)*e^(ln10*tan(πθ))
= ln10.πsec²(πθ)*10^(tan (πθ))

b) y= √[tln(t^4)]
= √[4t*ln(t)]
= 2√[t*ln(t)]
= 2u^½ where u = t*ln(t)
dy/dt = dy/du * du/dt
= u^-½ * (ln(t) + 1)
= (ln(t) + 1)/√[tln(t)]

c) y = arctan(arcsin (x^½))
= arctan u where u = arcsin v where v = x^½
dy/dx = dy /du * du/dv * dv/dx
= 1/(1+ u²) * 1/√[1 - v²] * ½x^(-½)
= 1/[2√(1 - x)*(1 + (arcsin (x^½))²)]

d)xe^y = y - 1
e^y + xe^y * dy/dx = dy/dx
So dy/dy - xe^y * dy/dx = e^y
ie dy/dx (1 - xe^y) = e^y
dy/dx = e^y/ (1 - xe^y)
= 1/(e^(-y) - x)

Answer 2

You say "find the derrivative", but what you should be saying is: find the derrivative of y with respect to x. I hate to be pushy about minor details, but that's what mathematics is all about.

A)
y=10^(tan[pi*theta]) --in the future you will be more clear with your notation; I know this text window sucks

we know that d/dx a^u = (a^u)*(ln 10)*(du/dx) where u = 10^(tan[pi*theta]) and a = 10 --this is some thorem we learned in calc 1.

so y'={10^(tan[pi*theta])}*{ln 10}*[d/dx tan (pi*theta)]

we know that d/dx (tan u) = [(sec u)^2]*(du/d(theta)) where u=pi*theta --I wish I had the theta symbol, some integral notation would be nice to...

so y'={10^(tan[pi*theta])}*{ln 10}*[(sec pi*theta)^2]*pi --final answer

see, this is easy, just a few applications of the chain rule, and a thorem

B)
y = [t*ln (t^4)]^(1/2) --find the derrivative of y with respect to t

first we look at this problem as y = u^(1/2), where u = t*ln(t^4)

we know that dy/dt = (1/2)*[u^(-1/2)]*du/dt

so we have (1/2)*{[t*ln (t^4)]^(-1/2)}*du/dt

du/dt = t*[1/(t^4)]*(4*(t^3))] + [ln (t^4)] --mulitple chain rules, the product thorem, and the d/dt ln (u) = (1/u)*(du/dt) thorem

putting it all together

y' = [(1/2)*{[t*ln (t^4)]^(-1/2)}]*{t*[1/(t^4)]*(4*(t^3))] + [ln (t^4)]}

after we multiply it out, cancel, and algebraicly simplify; we end up with:

y' = (1/2)*[{ln (t^4)}+4]*[t*(ln (t^4))]^(1/2) --final answer; it could be written in a much prettier way on paper

C)
y = arctan(arcsin [x^(1/2)]) --find the derrivative of y with respect to x

we know from calc 1 that dy/dx arctan (u) = [1/((u^2)+1)]*(du/dx) in this case u = arcsin (x^(1/2))

once again, we apply the chain rule

y' = [1/(({arcsin (x^(1/2))}^2)+1)]*(du/dx)

we know from calc 1 that dy/dx arcsin (v) = [(1-(v^2))^(-1/2)]*(dv/dx)

v=x^1/2
du/dx = [(1-x)^(-1/2)]*(1/2)*(x^(-1/2))

now,

y' = [1/(({arcsin ((x)^(1/2))}^2)+1)]*{[(1-x)^(-1/2)]*(1/2)*(x^(-1/2))}
--final answer

I don't believe it will break down any further

D)
x*(e^y) = y-1 --find the derrivative of y with respect to x

here we will need to use implicit differentation (calc 1)

differentiate both side with respect to x

(e^y)+x*(e^y)*(dy/dx) = (dy/dx) --we used the product rule on the left side of the equation

now we solve for (dy/dx), a little algebra

(e^y)+x*(e^y)*(dy/dx)-(dy/dx)=0

factor out (dy/dx)

(e^y)+[x*(e^y)-1]*(dy/dx)=0

[x*(e^y)-1]*(dy/dx)=-(e^y) --subtract (e^y) from both sides and then divide by [x*(e^y)-1]

dy/dx = (-e^y)/[x*(e^y)-1] --final answer

good question, that was fun

Answer 3

Let's try b)

First apply the power rule: 4t^3
Now we have a product: t ln (t^4), so we get

t d[ln (t^4)] + ln (t^4) = [t / (t^4)] + ln (t^4) = (1 / t^3) + ln (t^4)

Now we apply the power rule again:

(1/2) * [t ln (t^4)]^-1/2

Putting them all together, we get

4t^3 * (1/2) * [t ln (t^4)]^-1/2 * [(1 / t^3) + ln (t^4)]
2t^3 * [t ln (t^4)]^-1/2 * ([1 / t^3) + ln (t^4)]

In a). what does Pi Theta mean?

Answer 4

ok. on the 1st you make the main of the quotient rule z' = ](a million+x^2)*0 -2*2x]/(a million+x^2)^2 = -4x/(a million+x^2)^2 z" follows the comparable technique. [(a million+x^2)^2*-4 - (-4x)*2*(a million+x^2)*2x]/(a million+x^2)^4. The numerator may be able to be factored to (a million+x^2)*[-4(a million+x^2) +16x]/(a million+x^2)^4. y' = 4*[(3x^2+4)^3]*6x = 24x*(3x^2+4)^3. Now use the product rule and chain rule to get y'' y" = 24x*[3*(3x^2+4)^2]*6x + 24*(3x^2+4)^3 = {24*(3x^2+4)^2}*[21x^2+4] that's tedious to declare the least.

Answer 5

a) y = e^[(ln10)tan(PiTheta)]
dy/dTheta = e^[(ln10)tan(PiTheta)](ln10)sec^2(PiTheta)(Pi)
dy/dTheta=[Pi(ln10)]sec^2(PiTheta)10^(tanPiTheta)

b) y = [t ln (t^4)]^(1/2) = [4t ln(t)]^(1/2) = 2 [t ln(t)]^(1/2)
note that dy/dt t ln(t) = 1 + ln(t)
dy/dt = (2)(1/2)[t ln(t)]^(-1/2)[1 + ln(t)]
dy/dt = [1 + ln(t)]/sqrt[t ln(t)]

c) Note that dy/dx atan(x) = 1/(1+x^2)
and that dy/dx asin(x) = 1/sqrt(1-x^2)
dy/dx atan(asin(x^.5)) = [1/(1+asin(x^.5)^2)][1/sqrt(1-x)](.5)x^(-.5)
dy/dx = 1/[2(x^.5)sqrt(1-x)(1+asin(x^.5)^2)]

d) x(e^y)dy+(e^y)dx = dy
(e^y)dx = dy[1-x(e^y)] = dy[1-(y-1)] = -ydy
dy/dx = - (e^y)/y or -x(e^y)/(xy) which is (1-y)/(xy)