How do you find the derivative of #y= 10^(1-x^2)# ?

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Answer 1 · 2014-07-26T04:05:23

The answer is : #y'=-2ln10*x*10^(1-x^2)#

Solution:

let's #y=a^f(x)#, where a is some constant and #f(x)# is function of #x#

then, #y'= a^f(x)lna*f'(x)#

Similarly, if we follow this chain rule, we will get

#y'=10^(1-x^2)ln10*(-2x)#

After rearranging, we get

#y'=-2ln10*x*10^(1-x^2)#