Use first principles to find the gradient of y=tanh(x)?

1 Answer
May 24, 2018

Given y=f(x), f'(x)=lim_(hto0)(f(x+h)-f(x))/h

f'(x)=lim_(hto0)(tanh(x+h)-tan(x))/h

f'(x)=lim_(hto0)((tanh(x)+tanh(h))/(1+tanh(x)tanh(h))-tan(x))/h

f'(x)=lim_(hto0)((tanh(x)+tanh(h))/(1+tanh(x)tanh(h))-(tanh(x)+tanh(h)tanh^2(x))/(1+tanh(x)tanh(h)))/h

f'(x)=lim_(hto0)((tanh(x)+tanh(h)-tanh(x)-tanh(h)tanh^2(x))/(1+tanh(x)tanh(h)))/h

f'(x)=lim_(hto0)(tanh(x)+tanh(h)-tanh(x)-tanh(h)tanh^2(x))/(h(1+tanh(x)tanh(h)))

f'(x)=lim_(hto0)(tanh(h)-tanh(h)tanh^2(x))/(h(1+tanh(x)tanh(h)))

f'(x)=lim_(hto0)(tanh(h)(1-tanh^2(x)))/(h(1+tanh(x)tanh(h)))

f'(x)=lim_(hto0)(tanh(h)sech^2(x))/(h(1+tanh(x)tanh(h)))

f'(x)=lim_(hto0)(sinh(h)sech^2(x))/(hcosh(h)(1+tanh(x)tanh(h)))

f'(x)=lim_(hto0)sinh(h)/h*lim_(hto0)sech^2(x)/(cosh(h)(1+tanh(x)tanh(h)))

f'(x)=1*sech^2(x)/(cosh(0)(1+tanh(x)tanh(0)))

f'(x)=1*sech^2(x)/(1(1+0))

f'(x)=sech^2(x)