Question #edef2

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Oct 12, 2017

#dy/dx=y(cosx*log(2^x+7)+((2^xln(2))/2^x+7)*sinx)#

Explanation:

We have a function #y=(2^x+7)^sinx#
Here we'll take #log # on both sides of the equation because the function is in the form of a variable to the power of a variable.

#logy=log(2^x+7)^sinx#

Using the logarithmic identity, #log_ba^n=nlog_ba# we get

#log y=sinx*log(2^x+7)#

Now we will differentiate both sides with respect to #x# using the chain rule and the product rule.

#d/dx(logy)=d/dx(sinx*log(2^x+7))#

As we know the derivative of #a^x# is #a^xln(a)#

#1/ydy/dx=cosx*log(2^x+7)+((2^x*ln(2))/(2^x+7))*sinx#

Therefore

#dy/dx=y(cosx*log(2^x+7)+((2^xln(2))/2^x+7)*sinx)#

You can put the value of #y# in there if you want.

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