Question

what is the derivatives of sin6x.cos4x

Answer 1

Answer:

6cos4xcos6x - 4sin6xsin4x

Answer 2

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

We have to find derivative of sin6x.cos4x

y= sin6x.cos4x

differentiating w.r.t.x

=>dy/dx=(sin6x)'cos4x+(cos4x)'(sin6x)

product rule : u'v+v'u

=>dy/dx=cos6x(6x)'cos4x-sin4x(4x)'sin6x

=>Here, derivatives will be apply two times e.g., sin6x so,first find derivatives of sin6x whole term then find derivatives of its angle say 6x which will be 6.

=>similarly,in cos4x first we find derivatives of cos4x whole then find derivatives of 4x which is angle.

=>dy/dx= 6cos6xcos4x-4sin4xsin6x

Extra information=>

Derivatives of various terms and trigonometric

1) (xⁿ)'= nxⁿ⁻¹

2)(constant)'=0

3)(x)'=1

4)(1/x)'=-1/x²

5)(sinx)'=cosx

6)(cosx)'=-sinx

7)(tanx)'=sec²x

8)(cotx)'=-cosec²x

9)(secx)'=secxtanx

10)(cosecx)'=-cosecxcotx

11)(eˣ)'=eˣ

12)(logx)'= 1/x