If f(x)=sin^5x, Find f'(x)
i don't know im doing right way
f'(x)= 5cosx * sin(x)
im guessing you meant sin(x)^5. in which case yeah you did it wrong. you derive (x)^k , k being a constant, like this k*(x)^(k-1)*x'
so f'(x)=5sin(x)^4*cos(x)
Assuming that you mean [sinx]^5, the derivative is
5(sinx)^4 * cosx
I used the
"function of a function" rule for derivatives
I agree with Bryant
my worksheet say sin^5 x. well thanks
To find the derivative of f(x) = sin^5(x), you can use the chain rule.
First, rewrite the function as f(x) = (sin(x))^5.
Next, apply the chain rule. The general form of the chain rule states that if you have a function g(h(x)), the derivative of g(h(x)) with respect to x is given by g'(h(x)) * h'(x). In this case, g(u) = u^5 and h(x) = sin(x).
To find g'(u), you can apply the power rule for differentiation. The power rule states that if you have a function u^n, the derivative with respect to u is given by n * u^(n-1). In this case, n = 5, so g'(u) = 5u^4.
To find h'(x), differentiate sin(x) with respect to x. The derivative of sin(x) is cos(x).
Now, use the chain rule formula to find the derivative of f(x):
f'(x) = g'(h(x)) * h'(x)
= 5(sin(x))^4 * cos(x)
= 5sin^4(x) * cos(x)
So, f'(x) = 5sin^4(x) * cos(x).