Find the derivative of the given function.f(x)=(x3+4x)sin−1xf(x)=(x^3+4x)\sin^{-1}xf(x)=(x3+4x)sin−1x

x 3 + 4 x 1 − x 2 + ( 3 x 2 + 4 ) sin ⁡ − 1 x \frac{x^3+4x}{\sqrt{1-x^2}}+\left(3x^2+4\right)\sin^{-1}x 1 − x 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> x 3 + 4 x + ( 3 x 2 + 4 ) sin − 1 x

Find the derivative of the given function. f ( x ) = ( x 3 + 4 x ) sin − 1 x f(x)=(x^3+4x)\sin^{-1}x f ( x ) = ( x 3 + 4 x ) sin − 1 x

Here, we wanna find the derivative of this function f of x. That is x cubed+4x times the inverse sine of x. So in finding the derivative here, f prime of x, we need to use the product rule since I have one function multiplying another function. Remember our memory tool for the product rule, of course, left d right plus a right d left. That's our left hand function. So here, x cubed plus 4x times the derivative of that right hand function. Now we just learned that the derivative of the inverse sine of x is going to be 1 over the square root of 1 minus x squared. Then continuing on with our product rule here, plus right d left, so that's the inverse sine of x, times the derivative of this x cubed plus 4 x, that is 3 x squared plus 4. Now I can write this a little bit nicer because I can go ahead and condense this all into one fraction. That is x cubed plus 4 x over the square root of 1 minus x squared. And then plus here, we typically like to have all of our x terms in the front. So this is 3x squared plus 4 times the inverse sine of x. And this is my full derivative here, f prime of x. So as we learn more and more rules for finding derivatives, we can continue to combine them with all of the rules that we've already learned. Like here, we used our product rule along with our new derivative rule for finding the derivative of inverse sine. If you have any questions, let us know, and I'll see you in the next one.

Find the derivative of the given function. f ( x ) = ( x 3 + 4 x ) sin ⁡ − 1 x f(x)=(x^3+4x)\sin^{-1}x f ( x ) = ( x 3 + 4 x ) sin − 1 x

x 3 + 4 x 1 − x 2 + ( 3 x 2 + 4 ) sin ⁡ − 1 x \frac{x^3+4x}{\sqrt{1-x^2}}+\left(3x^2+4\right)\sin^{-1}x 1 − x 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> x 3 + 4 x + ( 3 x 2 + 4 ) sin − 1 x

3 x 2 + 4 1 − x 2 \frac{3x^2+4}{\sqrt{1-x^2}} 1 − x 2 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> 3 x 2 + 4

( 3 x 2 + 4 ) sin ⁡ − 1 ⁡ x + ( x 3 + 4 x ) cos ⁡ − 1 x (3x^2+4)\sin^{-1}⁡x+(x^3+4x)\cos^{-1}x ( 3 x 2 + 4 ) sin − 1 ⁡ x + ( x 3 + 4 x ) cos − 1 x

( 3 x 2 + 4 ) sin ⁡ − 1 ⁡ x − ( x 3 + 4 x ) cos ⁡ − 1 x (3x^2+4)\sin^{-1}⁡x-(x^3+4x)\cos^{-1}x ( 3 x 2 + 4 ) sin − 1 ⁡ x − ( x 3 + 4 x ) cos − 1 x

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Inverse Trigonometric Functions

Calculus

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Inverse Trigonometric Functions

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Find the derivative of the given function.
$f(x)=(x^3+4x)\sin^{-1}x$

$\frac{x^3+4x}{\sqrt{1-x^2}}+\left(3x^2+4\right)\sin^{-1}x$

$\frac{3x^2+4}{\sqrt{1-x^2}}$

$(3x^2+4)\sin^{-1}x+(x^3+4x)\cos^{-1}x$

$(3x^2+4)\sin^{-1}x-(x^3+4x)\cos^{-1}x$

Verified step by step guidance

Identify the function f(x) = (x^3 + 4x) \sin^{-1}x. This is a product of two functions: u(x) = x^3 + 4x and v(x) = \sin^{-1}x.

Apply the product rule for differentiation, which states that if you have a function f(x) = u(x)v(x), then its derivative f'(x) = u'(x)v(x) + u(x)v'(x).

Differentiate u(x) = x^3 + 4x. The derivative u'(x) is found by applying the power rule: u'(x) = 3x^2 + 4.

Differentiate v(x) = \sin^{-1}x. The derivative v'(x) is \frac{1}{\sqrt{1-x^2}}, which is the standard derivative of the inverse sine function.

Substitute u'(x), v(x), u(x), and v'(x) into the product rule formula: f'(x) = (3x^2 + 4)\sin^{-1}x + (x^3 + 4x)\frac{1}{\sqrt{1-x^2}}.

7:26m

Watch next

Master Derivatives of Inverse Sine & Inverse Cosine with a bite sized video explanation from Patrick

Start learning

Derivatives of Inverse Trigonometric Functions practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice