Struggling with Derivatives

Question

Hi everyone,

I'm currently taking Calculus I and I'm really hitting a wall with derivatives. I understand the basic concept of the slope of a tangent line and the power rule, but when things get even slightly more complicated, I get completely lost.

For example, problems involving the product rule, quotient rule, and especially the chain rule are just making my head spin. I've watched countless videos and read through my textbook multiple times, but it's still not clicking.

I'm trying to work through some practice problems, and I keep getting the wrong answers block blast. It's really frustrating because I feel like I should be able to understand this, but something just isn't connecting.

Has anyone else struggled with derivatives? If so, what resources or strategies did you find helpful? Are there any common mistakes I should be looking out for?

Answer 1

In calculus you must have perfect algebra.,,,,by the way, I use * as a multiplication sign.

It is simple to remember product rule

if h(x)=f(x)*g(x)

example h(x)=x²*sin(x)

then dh/dx=h')(x)=f'(x)*g(x)+f(x)*g'(x)

h'(x)=2x*sin(x)+x²*cos(x)=x*(2sin(x)+x*cos(x)

you have to know how to find f'(x) and g'(x), multiply them with g(x), f(x), and simplify with algebra

less simple is quotient rule. I use the product rule and calculus to remember

have h(x)=f(x)/g(x),,,same as h(x)=f(x)*[1/g(x)]=f(x)*[g(x)]^-1

example h(x)=x²/sin(x)=x²*[sin(x)]^-1

then h'(x)=f'(x)*[1/g(x)]+f(x)*[-g(x)]^-2*g'(x)],,,,better written as h'(x)=[f'(x)*g(x)-f(x)*g'(x)]/[g(x)]²

h'(x)=2x/sin(x)+[x²]*[-sin(x)]^-2*cos(x)=[2x*sin(x)-x²cos(x)]/sin²(x)

it is often tough algebra to get f'(x) and g'(x) and multiply and divide with f(x) and g(x)

properly and simplify with no mistakes.

here we used product rule and chain rule for derivative of [g(x)]^-1,,,,

lets think about chain rule for q(x)=[g(x)]^-1

you know if p(x)=xⁿ then p'(x)=n*x^n-1*(dx/dx),,,,,dx/dx=1

well if q(x)=[g(x)]^-1 then q'(x)=-1*[g(x)]^-2*g'(x)

example q(x)=1/sin(x)=[sin(x)]^-1

q'(x)=-1*[sin(x)]^-2*cos(x)=cos(x)/sin²(x)

we would say q(x)=p(g(x)) and q'(x)=dp/dx=dp/dg*dg/dx,,,,dg/dx=g'(x)

when you have calculus rules well known, and are meticulous with your

algebra (and geometry and trigonometry) calculus can be fun and useful.

a lot of statistics work is based on calculus. The continuous probability density function

is integrated over a range to determine probability.