Question

derivative of sin^2(x)/x

Answer 1

Apply quotient rule. Do you know it?

Answer 2

No :(

Answer 3

Actual question is F(t)=sin^2(x)/x then what is F'(t)?

Answer 4

\[\frac{ vdu-udv }{ v^2 }\] That what you are going to apply. Do you understand this pattern?

Answer 5

F(t)=sin^2(x)/x = sin^2(x) * x^(-1)
Now, use product rule ( product rule and quotient rule are somewhat same)
Product rule is : d(f(x) * g(x))/dx = f(x) * g'(x) + f'(x) * g(x)

Answer 6

hmm i would not recommend that

Answer 7

what @satellite73

Answer 8

i would not recommend using the product rule
don't be scared of the quotient rule

Answer 9

\[\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}\] with
\[f(x)=\sin^2(x), f'(x)=2\sin(x)\cos(x), g(x)=x, g'(x)=1\]

Answer 10

Ya i do understand this formula @Yttrium
But then to i'm new for it.
So how to solve next

u=sin^2x and v=t

t(d/dt sin^2 t)-sin^2t(1)/t^2

right?? @Yttrium

derivative of sin^2 (x) will be what?

Answer 11

you need the chain rule for that

Answer 12

Now what is chain rule?? :o

Answer 13

you have a composite function
the square of the sine of \(x\)

Answer 14

\[\left(f\circ g\right)'=f'(g)g'\]

Answer 15

in this case \(f(x)=x^2\) and \(g(x)=\sin(x)\) so \(f\circ g(x)=f(\sin(x))=[\sin(x)]^2=\sin^2(x)\)

Answer 16

so what would be the final answer?

Answer 17

i wrote it above

Answer 18

the derivative of something squared is two times something, times the derivative of something
the derivative of sine squared is two times sine, times the derivative of sine
the derivative of sine squared, is two times sine, times cosine

Answer 19

Ohh that is so confusing but i understood. :) @Satellite73

So just check

F(t)= sin^2(x)/x

F'(t)={x[2(sin x)(cos x)]-sin^2 x}/x^2

Answer 20

@Swara , you got it right! Just continue it to arrive at an answer. :))