Question

If y=e^alogx+e^xloga then dy/dx=

Answer 1

Answer:

dy/dx = a.[e^(a-1)logx]+ (e^xloga).ln(a)

Step-by-step explanation:

Here, y=e^alogx+e^xloga

=> y= x^a + a^x

=>Let, taking both sides derivative...

dy/dx = (a.x^(a-1)) + (a^x . ln(a))

dy/dx = a.[e^(a-1)logx]+ (e^xloga).ln(a)