Ignatius Yeo's answer to QN's Junior College 2 H1 Maths Singapore question.
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The picture shows 2 methods of doing it - product rule and chain rule. For chain rule, let f(x) be x^2+2. Differentiate f(x) and multiply e^f(x) as shown. Hope it helps!
Date Posted:
4 years ago
Thanks, I get the second one!
But the first one, why is e^x^2 when differentiated is e^x^2(0)?
But the first one, why is e^x^2 when differentiated is e^x^2(0)?
Whenever you differentiate a constant value, it is 0. The first method is product rule. First term multiply by derivative of second term, and second term multiply derivative of first term. e^2 is a constant with no variables. Therefore d/dx e^2=0.
Product rule...
d/dx (e^(x²) e²)
= e^(x²) d/dx (e²) + (d/dx e^(x²) ) e²
= e^(x²) (0) + 2xe^(x²) (e²)
= 2xe²e^(x²)
= 2x e^(x²+2)
Remember that differentiating a constant gives 0
d/dx (e^(x²) e²)
= e^(x²) d/dx (e²) + (d/dx e^(x²) ) e²
= e^(x²) (0) + 2xe^(x²) (e²)
= 2xe²e^(x²)
= 2x e^(x²+2)
Remember that differentiating a constant gives 0
Another tip d/dx e^2 = d/dx e^2(x^0). Applying differentiation rule, d/dx (e^2(x^0))= 0*x^-1*e^2=0. Anything multiplies 0 is always 0. Hence, d/dx constant =0.
Thanks a lot!!
You're welcome.
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