If # y = (sinx)^(sinx) # then find #dy/dx#?

Question

Isabella Ackerman · Accepted Answer

#y'=sin(x)^sin(x)cos(x)*(ln(sin(x))+1)#

We want to find the derivative of

#y=(sin(x))^sin(x)#

Take the logarithm on both sides

#ln(y)=ln((sin(x))^sin(x))#

#ln(y)=sin(x)*ln(sin(x))#

Differentiate both sides using the product and chain rule

(Be aware of the implicit differentiation on the right side)

#y'*1/y=cos(x)ln(sin(x))+sin(x)cos(x)*1/sin(x)#

#y'=y*(cos(x)ln(sin(x))+sin(x)cos(x)*1/sin(x))#

#y'=y*(cos(x)ln(sin(x))+cos(x))#

#y'=y*cos(x)*(ln(sin(x))+1)#

Substitute #y=(sin(x))^sin(x)#

#y'=(sin(x))^sin(x)cos(x)*(ln(sin(x))+1)#

Scarlett Allison · Answer

#dy/dx= (sinx)^sinx (cosx+ cosx log sinx)# #y= (sinx)^sinx # Taking log on both sides we get , #log y = sinx log sinx # Differentiating both sides we get , #1/y*dy/dx= sinx * 1/sinx *cosx+ cosx log sinx#or #1/y*dy/dx= cancel(sinx) * 1/cancel(sinx) *cosx+ cosx log sinx# or #dy/dx= y (cosx+ cosx log sinx)# or #dy/dx= (sinx)^sinx (cosx+ cosx log sinx)# [Ans]

Elijah Abram · Answer

#sin^sin(x)(x)(cos(x)lnsin(x)cos(x))#

We have to differentiate #sin(x)^sin(x)#.

Or maybe, #d/dxsin^sin(x)(x)#.

Remember, #a^b=e^(b*lna)#.

So the above becomes #d/dxe^(sin(x)*lnsin(x)#

Apply the chain rule:

The rule states that #(df(u))/dx=(df)/(du)*(du)/dx#.

Here, #f=e^u# and #u=sin(x)*lnsin(x)#

The equation simplifies (or gets excruciatingly confusing) to

#d/(du)e^u*d/dxsin(x)*lnsin(x)#

The derivative of #e^x# is #e^x#. So now the equation is:

#e^u*d/dxsin(x)*lnsin(x)#

Now for the second half. Remember the product rule: #(f*g)'=f'g+fg'#.

So now the equation is:

#e^u*d/dx(sin(x))*lnsin(x)+d/dx(lnsin(x))*sin(x)#

The derivative of #sin(x)# is #cos(x)#.

So the first derivative is #cos(x)lnsin(x)#. The equation is now #e^ucos(x)lnsin(x)*d/dx(lnsin(x))*sin(x)#

Sadly, we must use the chain rule again. Here, I take it as the differentiation of #f(w)#.

#f=lnw#, and #w=sin(x)#

The derivative of #lnw# is #1/w#, and #sin(x)# is again #cos(x)#

We now have #cos(x)/w#. But remember, #w=sin(x)#.

So this becomes #cos(x)/sin(x)#, which simplifies to #cot(x)#.

Our equation is presently #e^ucos(x)lnsin(x)cot(x)sin(x)#.

But #cot(x)sin(x)=cos(x)/sin(x)*sin(x)=cos(x)#

We now have #e^ucos(x)lnsin(x)cos(x)#. We're not done yet, though.

#u=sin(x)lnsin(x)#.

The equation is #e^(sin(x)lnsin(x))cos(x)lnsin(x)cos(x)#.

But #e^(b*lna)=a^b#.

So now, the equation is #sin^sin(x)(x)(cos(x)lnsin(x)cos(x))#

This, (thank God), cannot be simplified further.

Cameron Alexander · Answer

# dy/dx = cosx \ (sinx)^(sinx) \ {1+ ln sinx }#

We have: # y = (sinx)^(sinx) # If we take Natural Logarithms, we have: # ln y = ln {(sinx)^(sinx) }# And using the properties of logarithms we have: # ln y = sinx \ ln sinx# We can now readily differentiate wrt #x# by applying the chain rule (or implicit differentiation the LHS and the chain rule and the product rule on the RHS: # 1/y \ dy/dx = (sinx)(1/sinx cosx) + (cosx)ln sinx # Which we can simplify: # 1/y \ dy/dx = cosx + cosx \ ln sinx # # :. dy/dx = y{cosx + cosx \ ln sinx }# # \ \ \ \ \ \ \ \ \ \ = ycosx{1+ ln sinx }# # \ \ \ \ \ \ \ \ \ \ = cosx \ (sinx)^(sinx) \ {1+ ln sinx }#

Christopher Agresta · Answer

undefined To find $ \frac{dy}{dx} $ for the given function $ y = (\sin x)^{\sin x} $, we will use logarithmic differentiation.

Taking natural logarithm on both sides:
$$ \ln y = \ln \left( (\sin x)^{\sin x} \right) $$
$$ \ln y = \sin x \cdot \ln(\sin x) $$

Now, differentiate implicitly with respect to $ x $:
$$ \frac{1}{y} \cdot \frac{dy}{dx} = \cos x \cdot \ln(\sin x) + \frac{\sin x}{\sin x} \cdot \frac{1}{\sin x} $$
$$ \frac{dy}{dx} = y \left( \cos x \cdot \ln(\sin x) + \cot x \right) $$

Substitute back $ y = (\sin x)^{\sin x} $ into the equation:
$$ \frac{dy}{dx} = (\sin x)^{\sin x} \left( \cos x \cdot \ln(\sin x) + \cot x \right) $$

Thus, $ \frac{dy}{dx} = (\sin x)^{\sin x} \left( \cos x \cdot \ln(\sin x) + \cot x \right) $.