Question

Differentiate with respect to x, a^x + x^e + e^a + e^√x

Answer 1

Answer:
(a^x)loga+(x^e)logx+(e^a)+[(e^√x)/(2√x)]
Step-by-step explanation:
d(a^x)/dx =(a^x).loga
d(x^e)/dx =(x^e).logx
d(e^a)/dx =(e^a).loge=e^a (because loge=1)
d(e^√x)/dx = (e^√x).(loge).[d(√x)/dx]
=(e^√x).[1/(2√x)]
=(e^√x)/(2√x)
So by adding all 4 differentiations, answer becomes,
d[f(x)]/dx=
(a^x)loga+(x^e)logx+(e^a)+[(e^√x)/(2√x)]