Question #82351

1 Answer
Feb 24, 2016

dy/dx=e^x(ln(sinx)+cotx)

Explanation:

To find the derivative of this expression, we'll use the product rule, which states:
d/dx(uv)=u'v+uv'
Where u and v are functions of some variable (usually x).

We can see that the problem satisfies the two basic requirements of the product rule: both e^x and ln(sinx) are functions of x, and they're being multiplied together. So we can let u=e^x and v=ln(sinx) and apply the rule. First, we'll have to find the derivatives of these two functions:
d/dx(e^x)=e^x

d/dx(ln(sinx))=cosx*1/sinx=cosx/sinx=cotx->using chain rule and lnx rule

Now we have u (e^x), u' (e^x), v (ln(sinx)), and v' (cotx). All that is left is to plug these guys into the formula and simplify:
dy/dx=(e^x)(ln(sinx))+(e^x)(cotx)
=e^x(ln(sinx)+cotx)

Remember that u and v don't have to be functions of x; they can be functions of anything as long as it's the same variable (as in 3z^2cosz).