# chain rule problems

), Solution 2 (more formal). Great problems for practicing these rules. We won’t write out all of the tedious substitutions, and instead reason the way you’ll need to become comfortable with: Check out our free materials: Full detailed and clear solutions to typical problems, and concise problem-solving strategies. Chain Rule problems Use the chain rule when the argument of the function you’re differentiating is more than a plain old x. Here are a few problems where we use the chain rule to find an equation of the tangent line to the graph $$f$$ at the given point. Section 3-9 : Chain Rule For problems 1 – 51 differentiate the given function. We’ll illustrate in the problems below. Because the argument of the sine function is something other than a plain old x, this is a chain rule problem. So when using the chain rule: Example $$\PageIndex{9}$$: Using the Chain Rule in a Velocity Problem. Solution 2 (more formal) . Then. The problem that many students have trouble with is trying to figure out which parts of the function are within other functions (i.e., in the above example, which part if g(x) and which part is h(x). It’s also one of the most important, and it’s used all the time, so make sure you don’t leave this section without a solid understanding. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$f\left( x \right) = {\left( {6{x^2} + 7x} \right)^4}$$, $$g\left( t \right) = {\left( {4{t^2} - 3t + 2} \right)^{ - 2}}$$, $$R\left( w \right) = \csc \left( {7w} \right)$$, $$G\left( x \right) = 2\sin \left( {3x + \tan \left( x \right)} \right)$$, $$h\left( u \right) = \tan \left( {4 + 10u} \right)$$, $$f\left( t \right) = 5 + {{\bf{e}}^{4t + {t^{\,7}}}}$$, $$g\left( x \right) = {{\bf{e}}^{1 - \cos \left( x \right)}}$$, $$u\left( t \right) = {\tan ^{ - 1}}\left( {3t - 1} \right)$$, $$F\left( y \right) = \ln \left( {1 - 5{y^2} + {y^3}} \right)$$, $$V\left( x \right) = \ln \left( {\sin \left( x \right) - \cot \left( x \right)} \right)$$, $$h\left( z \right) = \sin \left( {{z^6}} \right) + {\sin ^6}\left( z \right)$$, $$S\left( w \right) = \sqrt {7w} + {{\bf{e}}^{ - w}}$$, $$g\left( z \right) = 3{z^7} - \sin \left( {{z^2} + 6} \right)$$, $$f\left( x \right) = \ln \left( {\sin \left( x \right)} \right) - {\left( {{x^4} - 3x} \right)^{10}}$$, $$h\left( t \right) = {t^6}\,\sqrt {5{t^2} - t}$$, $$q\left( t \right) = {t^2}\ln \left( {{t^5}} \right)$$, $$g\left( w \right) = \cos \left( {3w} \right)\sec \left( {1 - w} \right)$$, $$\displaystyle y = \frac{{\sin \left( {3t} \right)}}{{1 + {t^2}}}$$, $$\displaystyle K\left( x \right) = \frac{{1 + {{\bf{e}}^{ - 2x}}}}{{x + \tan \left( {12x} \right)}}$$, $$f\left( x \right) = \cos \left( {{x^2}{{\bf{e}}^x}} \right)$$, $$z = \sqrt {5x + \tan \left( {4x} \right)}$$, $$f\left( t \right) = {\left( {{{\bf{e}}^{ - 6t}} + \sin \left( {2 - t} \right)} \right)^3}$$, $$g\left( x \right) = {\left( {\ln \left( {{x^2} + 1} \right) - {{\tan }^{ - 1}}\left( {6x} \right)} \right)^{10}}$$, $$h\left( z \right) = {\tan ^4}\left( {{z^2} + 1} \right)$$, $$f\left( x \right) = {\left( {\sqrt{{12x}} + {{\sin }^2}\left( {3x} \right)} \right)^{ - 1}}$$. Students will get to test their knowledge of the Chain Rule by identifying their race car's path to the finish line. Think something like: “The function is some stuff to the $-2$ power. For how much more time would … Example problem: Differentiate y = 2 cot x using the chain rule. On problems 1.) We have the outer function $f(u) = e^u$ and the inner function $u = g(x) = \sin x.$ Then $f'(u) = e^u,$ and $g'(x) = \cos x.$ Hence \begin{align*} f'(x) &= e^u \cdot \cos x \8px] &= e^{\sin x} \cdot \cos x \quad \cmark \end{align*}, Solution 2 (more formal). Suppose that a skydiver jumps from an aircraft. Thanks to all of you who support me on Patreon. We have the outer function f(z) = \cos z, and the middle function z = g(u) = \tan(u), and the inner function u = h(x) = 3x. Then f'(z) = -\sin z, and g'(u) = \sec^2 u, and h'(x) = 3. Hence: \begin{align*} f'(x) &= (-\sin z) \cdot (\sec^2 u) \cdot (3) \\[8px] We have the outer function f(u) = \sin u and the inner function u = g(x) = 2x. Then f'(u) = \cos u, and g'(x) = 2. Hence \begin{align*} f'(x) &= \cos u \cdot 2 \\[8px] The chain rule for functions of more than one variable involves the partial derivatives with respect to all the independent variables. In other words, we always use the quotient rule to take the derivative of rational functions, but sometimes we’ll need to apply chain rule as well when parts of that rational function require it. A garrison is provided with ration for 90 soldiers to last for 70 days. If you still don't know about the product rule, go inform yourself here: the product rule. Now, for the first of these we need to apply the product rule first: To find the derivative inside the parenthesis we need to apply the chain rule. Let f(x)=6x+3 and g(x)=−2x+5. find answers WITHOUT using the chain rule. Solution: The derivatives of f and g aref′(x)=6g′(x)=−2.According to the chain rule, h′(x)=f′(g(x))g′(x)=f′(−2x+5)(−2)=6(−2)=−12. Work from outside, in. Huge thumbs up, Thank you, Hemang! Solution 2 (more formal). 2 Chain rule for two sets of independent variables If u = u(x,y) and the two independent variables x,y are each a function of two new independent variables s,tthen we want relations between their partial derivatives. Note: You’d never actually write out “stuff = ….” Instead just hold in your head what that “stuff” is, and proceed to write down the required derivatives. The aim of this website is to help you compete for engineering places at top universities. This is the currently selected item. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}} Please read and accept our website Terms and Privacy Policy to post a comment. Solution 2 (more formal). We have $y = u^7$ and $u = x^2 +1.$ Then $\dfrac{dy}{du} = 7u^6,$ and $\dfrac{du}{dx} = 2x.$ Hence \begin{align*} \dfrac{dy}{dx} &= 7u^6 \cdot 2x \8px] Category Questions section with detailed description, explanation will help you to master the topic. And what the chain rule tells us is that this is going to be equal to the derivative of the outer function with respect to the inner function. We provide challenging problems that are similar in style to some interview questions. • Solution 3. This can be viewed as y = sin(u) with u = x2. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. That material is here. Let’s use the first form of the Chain rule above: \[\bbox[10px,border:2px dashed blue]{\begin{align*}\left[ f\Big(g(x)\Big)\right]’ &= f’\Big(g(x)\Big) \cdot g'(x) \\[5px]&=\text{[derivative of the outer function, evaluated at the inner function] } \\[5px]&\qquad \times \text{ [derivative of the inner function]}\end{align*}} As another example, e sin x is comprised of the inner function sin Assume that t seconds after his jump, his height above sea level in meters is given by g(t) = 4000 − 4.9t 2. Here’s a foolproof method: Imagine calculating the value of the function for a particular value of $x$ and identify the steps you would take, because you’ll always automatically start with the inner function and work your way out to the outer function. &= 3\tan^2 x \cdot \sec^2 x \quad \cmark \\[8px] Determine where $$V\left( z \right) = {z^4}{\left( {2z - 8} \right)^3}$$ is increasing and decreasing. 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