Question

Question #edef2

Answer 1

`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.