Hi
Can you explain why this answer is correct?
Differentiate wrt(x)
\(\displaystyle \L \b
y = \sqrt {e^x } = (e^x )^{{\textstyle{1 \over 2}}}\)
\(\displaystyle \L u = e^x {\rm }u' = e^x\)
\(\displaystyle {\rm using the chain rule, } \b y = u^{\frac{1}{2}} {\rm }y' = \frac{1}{2}u^{ - \frac{1}{2}}\)
\(\displaystyle {\rm substituting back , }\L \b y' = \frac{1}{2}(e^x )^{ - \frac{1}{2}}\)
which is the right answer, but I dont understand why this expression isnt multiplied by the differential of u, that is, e^x, inside the bracket, the way you would if you were using the chain rule with a non-exponential function? Is there a special rule for e functions or am I missing something else?
Thanks
Can you explain why this answer is correct?
Differentiate wrt(x)
\(\displaystyle \L \b
y = \sqrt {e^x } = (e^x )^{{\textstyle{1 \over 2}}}\)
\(\displaystyle \L u = e^x {\rm }u' = e^x\)
\(\displaystyle {\rm using the chain rule, } \b y = u^{\frac{1}{2}} {\rm }y' = \frac{1}{2}u^{ - \frac{1}{2}}\)
\(\displaystyle {\rm substituting back , }\L \b y' = \frac{1}{2}(e^x )^{ - \frac{1}{2}}\)
which is the right answer, but I dont understand why this expression isnt multiplied by the differential of u, that is, e^x, inside the bracket, the way you would if you were using the chain rule with a non-exponential function? Is there a special rule for e functions or am I missing something else?
Thanks
