### Derivative of e^x and a^x

 We know the formula for finding the value of ex ex = 1 + x + x2/2! + x3/3! + x4/4! + ... We know the formula for finding the derivative ```dy/dx = lim [f(x + δx) - f(x)]/δx δx→0 ``` Derivative of ex y = f(x) = ex = 1 + x + x2/2! + x3/3! + x4/4! + ... f(x + δx) = ex + δx Difference f(x + δx) - f(x) = ex + δx - ex = exeδx - ex = ex(eδx - 1) = ex(1 + δx + δx2/2! + δx3/3! + δx4/4! + ... - 1) = ex(δx + δx2/2! + δx3/3! + δx4/4! + ...) = ex(δx(1 + δx/2! + δx2/3! + δx3/4! + δx4/5! + ...)) Ratio [f(x + δx) - f(x)]/δx = ex(δx(1 + δx/2! + δx2/3! + δx3/4! + δx4/5! + ...))/δx = ex(1 + δx/2! + δx2/3! + δx3/4! + δx4/5! + ...) ```dy/dx = lim [f(x + δx) - f(x)]/δx δx→0 ``` When δx →0 then all the terms containing δx tends to zero and become negligible and dy/dx = ex Derivative of ax y = f(x) = ax = ex loge a f(x + δx) = e(x + δx)loge a Difference f(x + δx) - f(x) = e(x + δx)loge a - ex loge a = ex loge aeδx loge a - ex loge a = ex loge a(eδx loge a - 1) = ex loge a(1 + δx loge a + (δx loge a)2/2! + (δx loge a)3/3! + (δx loge a)4/4! + ... - 1) = ex loge a(δx loge a + (δx loge a)2/2! + (δx loge a)3/3! + (δx loge a)4/4! + ...) = ex loge a(δx loge a(1 + (δx loge a)/2! + (δx loge a)2/3! + (δx loge a)3/4! + (δx loge a)4/5! + ...)) Ratio [f(x + δx) - f(x)]/δx = ex loge a(δx loge a(1 + (δx loge a)/2! + (δx loge a)2/3! + (δx loge a)3/4! + (δx loge a)4/5! + ...))/δx = ex loge a(loge a(1 + (δx loge a)/2! + (δx loge a)2/3! + (δx loge a)3/4! + (δx loge a)4/5! + ...)) ```dy/dx = lim [f(x + δx) - f(x)]/δx δx→0 ``` When δx →0 then all the terms containing δx tends to zero and become negligible and dy/dx = ex loge aloge a dy/dx = ax loge a d(ex)/dx = ex and d(ax)/dx = ax loge a